Property of join

 Let $X$ be path-connected and $Y$ be arbitrary topological space. Then the join $X*Y$ is simply connected.

 

$\textbf{Proof}.$ We use Van Kampen style argument. Define $U=X*Y\setminus Y=X\times Y\times [0,1)/\sim,$ where $\sim: \ (x,y_1,0)\sim(x,y_2,0) $ for all $y_1,y_2\in Y$.  Thus $U$ is clearly open and deformation retracts onto $X\cong  X\times Y\times \{0\}/\sim.$ As $X$ is connected, so is $U$.

Now fix $x_0\in X$. Let $V_1$ be the cone with apex $x_0$ and base  $\{x_0\}\times Y\times\{1/2\}$. Also define $V_2:=X\times Y\times [1/2,1]/\sim$, where $\sim: \ (x_1,y,0)\sim(x_2,y,0) $ for all $x_1,x_2\in X.$ Set $V=V_1\cup V_2$. Observe that $V_2$ deformation retracts onto  $\{x_0\}\times Y\times\{1/2\}$ (collapse everything to $Y$ and then push it to $\{x_0\}\times Y\times\{1/2\}$). But this is the base of the cone $V_1$, and $V_1$ deformation retracts to $x_0$. Hence $V$ is contractible.

The set $V$ is not open, but $U\cup \text{int}V=X$. Moreover, $U\cap V=V_1\cup (X\times Y\times [1/2,1))$ is connected. Thus every loop in $X*Y$ based at $x_0$ is homotopy equivalent to finite product of loops at $x_0$, each belonging to $U$ or $V$. Observe that $X$ is nullhomotopic in $X*Y$ (use a cone with base $X$ and apex some point in $Y$), and hence $U$ is such. Thus the inclusion $U\hookrightarrow X*Y$ induces a trivial homomorphism $\pi_1(U)\to \pi_1(X)$. Moreover, $\pi_1(V)$ is already trivial. Thus each loop in $X*Y$ is homotopy equivalent to a constant loop.

Homotopy type of R^3\(circles)

$\mathbf{1}$. The spaces $X=\mathbb{R}^3\setminus S^1$ and $S^{1}\vee S^{2}$ are homotopy equivalent.

First we rewrite $X\cong S^3\setminus (\{*\}\cup S^1)$, as $S^3$ is the one-point compactification of $\mathbb{R}^3$. Now observe $S^3\setminus S^1$ is homeomorphic to the open solid torus, $S^1\times e^2$, where $e^2$ is the open unit $2$-disc. An explicit homemorphism is as follows : $$(x,y,z,t)\mapsto (x\sqrt{1-z^2-t^2},y\sqrt{1-z^2-t^2},z,t).$$

 Thus $\mathbb{R}^3\setminus S^1\cong S^1\times e^2\setminus \{*\}$. Here $S^1\times e^2$ is open $3$-manifold. Removing a point from open $3$-manifold $M$ is wedging $M$ with $S^2$ $-$ we may think of enclosing the point with $S^2$ and retracting the enclosed region to the sphere (or blowing up the point to a sphere); then we may push the sphere away from the manifold, to the boundary, forming the wedge product. Thus $\mathbb{R}^3\setminus S^1\cong S^1\times e^2\vee S^2\sim S^1\vee S^2$. 

 

$\mathbf{2}$. The space $X=\mathbb{R}^3\setminus M$, where $M$ is the union of two disjoint unlinked circles, is homotopy equivalent to $S^{1}\vee S^1\vee S^{2}\vee S^2$.

As before, $X\cong S^1\times e^2\setminus(\{*\}\cup S^1)$. Observe that the remaining circle to be removed is contractible in $S^1\times e^2$, i.e. it is not encircling the hole of the torus. Thus we may take a $2$-sphere which bounds a region containing the circle to be removed. This regions retracts to $S^2\vee S^1$, as in $\mathbf{1}$. Pushing it to the boundary of $S^1\times e^2$ we get a wedge with $S^2\vee S^1.$ Removing the point $*$ induces, as in $\mathbf{1}$, one more wedge, with $S^2$. Hence $X\sim (S^1\times e^2)\vee S^2\vee S^2\vee S^1$, which finishes the derivations as $S^1\times e^2\sim S^1$

 

$\mathbf{3}$. The space $X=\mathbb{R}^3\setminus M$, where $M$ is the union of two disjoint linked circles, is homotopy equivalent to $S^{2}\vee (S^1\times S^1)$.

As in $\mathbf{2}$, $X\cong S^1\times e^2\setminus(\{*\}\cup S^1)$. Observe that the remaining circle to be removed is not contractible in $S^1\times e^2$, i.e. it encircles the hole of the torus. Removing the point $*$ induces, as in $\mathbf{1}$, a wedge with $S^2$.  Hence $X\sim ((S^1\times e^2)\setminus S^1)\vee S^2$. But $(S^1\times e^2)\setminus S^1$ is a solid torus with core circle removed, which clearly deformation retracts to a torus $-$ $S^1\times S^1$.

 

 

 

No retraction

 Let $I=[0,1]$, $A=\{0\}\cup\{\frac{1}{n}\}_{n\ge 1}$. Prove that there is no retraction from $I\times I$ to $I\times \{0\}\cup A\times I$. 

OMOUS 2024

Let $A$ be a real matrix such that $A+A^2A^t+(A^t)^2=0$. Prove that $A=0$.

$\textbf{Solution}.$ Multiply the given equation by $A^t$ from the right to get $AA^t+A^2(A^2)^t+(A^t)^3=0$ and hence $(A^t)^3=-AA^t-A^2(A^2)^t$ . Observe that RHS is negative semi-definite symmetric matrix. Thus $(A^t)^3$ and hence $A^3$ is symmetric negative semi-definite.

Now multiply the given equation with $A$ from the left to get  $A^2+A^3A^t+A(A^t)^2=0$ and using $A^3=(A^t)^3$ we get $A^2+(A^t)^4+A(A^t)^2=0$. Transposing this equation we get $(A^t)^2+A^4+A^2 A^t=0$. Comparing with the original equation we get $A^4=A$. Thus $A$ is diagonalizable and $\lambda^4=\lambda$ for every characteristic root $\lambda$ of $A$. Thus  $\lambda=0$ or $\lambda^3=1$. But $A^3$ is negative semi-definite, so $\lambda^3\le 0$. Thus $\lambda=0$ and hence all roots of $A$ are $0$. Since $A$ is diagonalizable, $A=0$.

Truncated exponential series equation

$\textbf{Problem}.$ For a positive integer $n$ let $x_n$ be the unique positive real root of the equation

$$\sum_{k=0}^n \frac{x^k}{k!}=\frac{e^x}{2}.$$

Prove that $\displaystyle\lim_{n\to\infty}(x_n-n)=\frac{2}{3}.$

$\textbf{Comment.}$ The limit $\displaystyle\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ is well known to be $\displaystyle\frac{1}{2}$. It could be derived via nice probabilistic structure related to Poisson distribution. This limit hints that the value of the root $x_n$ should be close to $n$. 

$\textbf{Sketch of a solution of the problem}$. Defining 

$$g(x)=\frac{1}{n!}\int_0^x s^ne^{-s}\ ds-\frac{1}{2}$$ 

the original equation could be rewritten in the form $g(x)=0.$ One should observe that $g$ is monotonically increasing and the unique root is in the interval $(n,n+1)$. Then one uses one iteration of the Newton method initialized at $n$ to obtain an approximation of the solution $\tilde x_n$. Using the final asymptotic obtained at the end of the this answer and Stirling approximation, one obtains that $\displaystyle\lim_{n\to\infty}(\tilde x_n-n)=\frac{2}{3}.$ It remains to use the estimate of the error in the Newton method (one needs the first and second derivatives of $g$ here, which are easy to write), to see that $\lim_{n\to\infty}(x_n-\tilde x_n)=0$.

Proof of Gelfand-Phillips Theorem

$\textbf{Theorem.}$ Let $X$ be a Banach space and $A\subseteq X$. Prove that $A$ is precompact (in the norm) if and only if for every $w^*$-convergent to $\textbf{0}$ sequence $\{x_n^*\}_{n\ge 1}\subseteq X^*$ it holds that $\{x_n^*\}_{n\ge 1}$ converges uniformly to $\textbf{0}$ on  $A.$

 

$\textbf{Proof.}$ If $A$ is precompact the result follows from Arzela-Ascoli theorem (combined with the fact that $w^*$-convergent sequences are norm bounded).

Now we prove the reverse direction. 

First we proof the following characterization of compactness in normed spaces: $A$ is compact if and only if $A$ is bounded and for any $\varepsilon>0$ there exists a finite-dimensional space $F$ such that $A\subseteq F+\varepsilon\mathbf{B}.$

$\textit{Proof}.$ 

If $A$ is precompact, then for any $\varepsilon>0$ there exists $\{a_i\}_{i=1}^n\subseteq A$ such that $A\subseteq\bigcup_{i=1}^n\mathbf{B}_{\varepsilon}(a_i)$. In particular $A$ is bounded and
\[A\subseteq \text{span}(\{a_i\}_{i=1}^n)+\varepsilon\textbf{B}.\]
For  the reverse direction, let $\varepsilon>0$ and $F$ be finite-dimensional space such that $A\subseteq F+\frac{\varepsilon}{3}\mathbf{B}.$ For each $a\in A$ choose $f_a\in F$ with $\|a-f_a\|\le\varepsilon/3$. Since $A$ is bounded and $F$ is finite-dimensional, the set $\{f_a\}_{ a\in A}$ is precompact.
Thus there exists $\{f_{a_i}\}_{i=1}^n\subseteq F$ such that $ \{f_a\}_{a\in A}\subseteq \bigcup_{i=1}^n\textbf{B}_{\varepsilon/3}(f_{a_i})$.
Now let $a\in A$ be arbitrary. Then there exists $i\in \{1,2,\ldots,n\}$ such that $\|f_a-f_{a_i}\|<\varepsilon/3.$ Consequently
\[\|a-a_i\|\le\|a-f_a\|+\|f_a-f_{a_i}\|+\|f_{a_i}-a_i\|<\varepsilon,\]
hence $\{a_i\}_{i=1}^n\subseteq A$ is a finite $\varepsilon$-net for $A.$ $\square$

Now let $A$ be a set which is not precompact. If it is unbounded, the results follows easily. Assume otherwise. Thus there exists $\varepsilon>0$ such that for any finite-dimensional space $F$, there exists $a\in A$ with $\text{d}(a,F)\ge\varepsilon.$ Choose a sequence $\{x_i\}_{i\ge 1}\subseteq X$ which is dense in $X$. Let $F_n:=\text{span}(\{x_i\}_{i=1}^n).$ Choose $a_n\in A$ with $\text{d}(a_n,F_n)\ge\varepsilon.$ Using Hahn-Banach construct $x_n^*\in X^*$ such that $\langle x_n^*,a_n\rangle\ge\varepsilon$, $\|x_n^*\|=1$ and $F_n\subseteq\text{ker}(x_n^*)$.
Thus for every $n$ \[\sup_{a\in A}|\langle x_n^*,a\rangle|\ge\varepsilon,\]
hence $\{x_n^*\}_{n\ge 1}$ does not converge uniformly to $\mathbf{0}$ on $A$. On the other hand $\bigcup_{n\ge 1}F_n$ is dense in $X$. Fix $x\in X$ and let $\delta>0$ be arbitrary. Then there exists $m$ such that for some $z_m\in F_m$ it holds that $\|z_m-x\|<\delta$. For $n\ge m$ it holds that $\langle x_n^*,z_m\rangle=0$, hence
\[|\langle x_n^*,x\rangle|= |\langle x_n^*,z_m-x\rangle|\le \|x_n^*\|\|z_m-x\|<\delta.\]
Consequently $\{x_n^*\}_{n\ge 1}$ is weak$^*$ convergent to $\mathbf{0}.$

Closed orbits are finite

 Let $f:[a,b]\to [a,b]$ be continuous function and $x_0\in [a,b]$. Consider the sequence $X=\{x_n\}_{n\ge 0}$ defined by $x_n=f(x_{n-1})$ for $n\ge 1$. Assume that $X$ is a closed set. Prove that it consists of finitely many elements. (The problem is taken from IMC 2002. The set $X$ is called the orbit of $f$ starting from $x_0$)

$\textbf{Proof.}$ Let us assume that $X$ is infinite. So it has a limit point, which should belong to the set (since it is closed). Let $$n_0:=\min\{n\ | \ x_n \text{ is a limit point of } X \}.$$ Then define $\hat X=\{x_n\}_{n\ge n_0+1}$. Thus $\hat X\cup \{x_{n_0}\}$ is compact and $f:\hat X\cup\{x_{n_0}\}\to \hat X$ is a continuous function. Since the image of compact sets under continuous functions are compact set, we must have that $\hat X$ is compact as well. But it is clearly not closed, since $x_{n_0}$ is a limit point $\hat X$ and does not belong to $\hat X$.

There is no lowest rate of convergence

 Let $\{a_n\}_{n\ge 1}$ be a sequence of positive numbers such that 

$$\sum_{n=1}^{\infty} a_n$$ is convergent. Then there exists a sequence $\{b_n\}_{n\ge 1}$ with $\displaystyle \lim_{n\to \infty}\frac{b_n}{a_n}=\infty$ and such that 

$$\sum_{n=1}^{\infty} b_n$$ is also convergent.

First proof (a la Calculus I, based on ideas of Zhivko Petrov). Consider $$S_n=\sum_{k=n}^{\infty}a_k.$$ Clearly $S_n\to 0$. Define $$b_n=\sqrt{S_n}-\sqrt{S_{n+1}}.$$

Clearly $b_n>0$ for all $n$ and $b_n\to 0$. Moreover 

$$\sum_{n=1}^\infty b_n=\sqrt{S_1},$$ hence is convergent.

On the other hand $$\lim_{n\to\infty}\frac{b_n}{a_n}=\lim_{n\to\infty}\frac{1}{\sqrt{S_n}+\sqrt{S_{n+1}}}=\infty.$$

 

Second proof. (Based on Folland's book) Assume on the contrary that there is lowest decaying sequence, i.e. 

for some sequence $a=\{a_n\}_{n\ge 1}$ with convergent series, holds that for any sequence of positive numbers $\{b_n\}_{n\ge 1}$, the series $$\sum_{n=1}^\infty a_nb_n$$ is convergent if and only if $\{b_n\}_{n\ge 1}$ is bounded (the if part holds for any sequence $\{a_n\}_{n\ge 1}$ with convergent series).

This implies that the linear operator $T:\ell_{\infty}\to\ell_1$ defined by 

$$T(c_1,c_2,\ldots)=(a_1c_1,a_2c_2,\ldots)$$ 

is not only well defined but also surjective ( due to the "only if" part above). Moreover the map is injective, since $a$ has only nonzero terms. The map is clearly bounded (by the norm of $a$), so the bounded inverse mapping theorem implies that the inverse $T^{-1}:\ell_1\to\ell_\infty$ is also bounded, hence continuous operator. Now it remains to observe that the image of $c_{00}$ under $T^{-1}$ is again $c_{00}$. However, $c_{00}$ is dense in $\ell_1$ and its image under continuous map should be dense in $\ell_\infty$, but the image (which is again $c_{00}$) is clearly not dense in $\ell_\infty$

Two functional inequalities

$\textbf{Problem 1.}$ Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x$ and $y$ holds

$$f(x)f(x+y)\ge f(x)^2+xy.$$

$\textbf{Problem 2.}$ Find all differentiable functions $f:\mathbb{R}\to\mathbb{R}$ with $f(0)=0,\ f(1)=1$ and such that for all $x$ and $y$ holds

$$f(x+y)\ge 2022^xf(x)+f(y).$$

Haudorff dimension and Liouville numbers

In this post we present a proof of the fact the the set of Liouville numbers contained in $[0,1]$ has Hausdorff dimension $0$. 

Why is this result interesting, apart from finding explicitly the Hausdorff dimension of a particular set? Regarding Lebesgue measure, it is not obvious whether there exists an uncountable set with measure zero. Lioville numbers serve as such example. However, Liouville numbers even serve as an example of an uncountable set with Hausdorff dimension zero, and this is much stronger. That is because every set of dimension zero has measure zero, whereas for example, another common reference of an uncountable set with measure $0$, the Cantor set, has positive Hausdorff dimension ($\ln 2/\ln 3$). Thus the set of Lioville numbers, is in a sense, way smaller than the Cantor set. One could also say that Liouville numbers are example of the smallest possible uncountable sets among the sets with measure zero. You can look here, for another construction of uncountable sets with dimension zero. Another interesting aspect of Liouville numbers is that they are $G_\delta$-dense in $[0,1]$ (which is a property stronger than uncountability). Hence Liouville numbers are big in sense of cardinality and topology, yet among the smallest with respect to measure and dimension.

Preliminary notions. We recall the definition of Hausdorff dimension.
For a subset $U$ of a metric space (e.g. real line), define $\operatorname{diam}(U)$ to be the diameter of $U$, i.e. the supremum of distances between any two elements of $U$. If $U$ would be an interval, that is just its length. Now for a set $X$, for $d\ge 0$ (which would take the role of a dimension) and $\delta>0$ define
$$H_\delta^d(X)=\inf\left \{\sum_{i=1}^\infty (\operatorname{diam} U_i)^d: X\subseteq\bigcup_{i=1}^\infty U_i,\ \operatorname{diam} U_i<\delta\right \}.$$
One observes that this is nonincreasing as a function of $\delta$ (smaller $\delta$ - less covers $\{U_n\}_{n\ge 1}$). Thus we define $$\displaystyle\mathcal{H}^d(X)=\sup_{\delta>0}H_\delta^d(X)=\lim_{\delta\to 0}H_\delta^d(X).$$ This turns out to be a measure (defined at least on the Borel $\sigma$-algebra (the one containing all open sets)); it is called $d$-dimensional Hausdorff measure. One may observe that $\mathcal{H}^d(X)$ is nonincreasing as a function of $s$ with values in $[0,\infty]$, and may obtain at most one finite nonzero value (consider for example subset of $[0,1]$).
Define the Hausdorff dimension as the $d$ for which we switch from measure $\infty$ to measure $0$, i.e.
$$\dim_{\operatorname{H}}{(X)}=\inf\{d\ge 0: \mathcal{H}^d(X)=0\}.$$

To the problem. We want to show that the set of Liouville numbers in $[0,1]$ has dimension $0$. Denote this set by $L$. Recall that by definition  $x\in L $ iff for each positive integer $n$ there exist infinitely many positive integers $p,q$, such that $$\left|x-\frac{p}{q}\right|<\frac{1}{q^n}.$$
Pay attention, the usual definition requires existence of at least one such pair $p,q$; it is easy to observe that if there exists at least one pair for each $n$, then there exist infinitely many pairs for each $n$.

So let us unpack what it means for a set to have dimension $0$. Going back to the above definitions (in reverse order) we see that it suffices to show that for any $s>0$ holds $\mathcal{H}^d(X)=0$. This in turn reduces to showing that for $\delta>0$ holds $H_\delta^d(X)=0$. This means that
$$\inf\left \{\sum_{i=1}^\infty (\operatorname{diam} U_i)^d: X\subseteq\bigcup_{i=1}^\infty U_i,\ \operatorname{diam} U_i<\delta\right \}=0$$
Finally, unwrapping the infimum we may summarise the task to be done as follows:

 For any $d>0$, $\delta>0$ and $\varepsilon>0$ there exists a countable collection $\{U_i\}_{i\ge 1}$  such that: $\displaystyle X\subset \bigcup_{i\ge 1}U_i$, $\operatorname{diam}U_i<\delta$ for all $i$ and
$$ \sum_{i=1}^\infty (\operatorname{diam} U_i)^d<\varepsilon.$$ So fix $d>0$, $\delta>0$ and $\varepsilon>0$. We may assume $d<1$. Fix $\displaystyle n>\frac{3}{d},\ q_0>\max\left\{\frac{2}{\delta},\frac{2}{\varepsilon}\right\}$ and consider
$$I_{p,q}:=\left(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\right).$$ One observes that (recalling the modified, still equivalent defintion, we stated) $$L\subset \bigcup_{q>q_0}\bigcup_{p=1}^{q-1}I_{p,q}$$ Moreover $\displaystyle\operatorname{diam}I_{p,q}=\frac{2}{q^n}<\frac{2}{q_0}<\delta.$ Finally
 $$\sum_{q>q_0}\sum_{p=1}^{q-1}(\operatorname{diam}I_{p,q})^d\le\sum_{q>q_0}\frac{2^d}{q^{nd}}q\le 2\sum_{q>q_0}\frac{1}{q^{nd-1}}\le 2\sum_{q>q_0}\frac{1}{q^{2}}\le \frac{2}{q_0}<\varepsilon.$$ In the above we used the inequality $$\sum_{k>n}\frac{1}{k^2}\le \frac{1}{n}$$ which could be proven, for example, by estimating the sum from above by $\displaystyle \int_n^\infty\frac{1}{x^2}dx$.

Thus the proof is finished.

You can also check the wikipedia articles for further information on Hausdorff dimension and Liouville numbers.

Problem proposed by prof. Gadjev

$\textbf{Problem.}$

 Let $f:[0,1]\to\mathbb{R}$ be a continuous function.

Find the limit 

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right).$$

 

$\textbf{Solution.}$ 

Fix $\varepsilon>0$. We aim to show that for large enough $n$  

$$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|<3\varepsilon$$

 which would mean that the limit is $f(1).$

Let $M$ be an upper bound for $|f|$ on $[0,1]$, i.e. $|f(x)|\le M$ for all $x\in [0,1].$

 For all $n\ge 1$ define $$k(n)=\left\lfloor\frac{n}{n^{1/\sqrt{\ln n}}}\right\rfloor.$$

The number is chosen in such a way that $$\lim_{n\to\infty}\frac{k(n)}{n}=0 \ \ \& \ \  \lim_{n\to\infty}\frac{\ln(k(n))}{\ln n}=1.$$

 Split the sum in two:

$$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|\le \left|\frac{1}{n}\sum_{k=1}^{k(n)} f\left(\frac{\ln k}{\ln n }\right)\right|+\left|\frac{1}{n}\sum_{k=k(n)+1}^{n} f\left(\frac{\ln k}{\ln n }\right)-f(1)\right| \ \ \ (*)$$

 Take $n_1$ such that for $n>n_1$ holds $\displaystyle \frac{k(n)}{n}<\frac{\varepsilon}{M}$. For the first summand on the right, using the triangle inequality, we obtain

$$\left|\frac{1}{n}\sum_{k=1}^{k(n)} f\left(\frac{\ln k}{\ln n }\right)\right|\le \frac{1}{n}\sum_{k=1}^{k(n)} \left|f\left(\frac{\ln k}{\ln n }\right)\right|\le \frac{Mk(n)}{n}<\varepsilon$$

when $n>n_1$. 

Now take $\delta>0$ such that when $|x-1|<\delta$ holds $|f(x)-f(1)|<\varepsilon$. Take $n_2$ such that for $n>n_2$ holds $\displaystyle \left|\frac{\ln(k(n))}{\ln n}-1\right|<\delta.$ Now for $n>\max\{n_1,n_2\}$ we obtain 

 $$\left|\frac{1}{n}\sum_{k=k(n)+1}^{n} f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|\le \sum_{k=k(n)+1}^{n} \frac{1}{n}\left|f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|+\left|\frac{k(n)}{n}f(1)\right|<\frac{n-k(n)}{n}\varepsilon+\varepsilon\le 2\varepsilon$$

 Thus for $n>\max\{n_1,n_2\}$ from $(*)$ we obtain 

$$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|\le \left|\frac{1}{n}\sum_{k=1}^{k(n)} f\left(\frac{\ln k}{\ln n }\right)\right|+\left|\frac{1}{n}\sum_{k=k(n)+1}^{n} f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|<\varepsilon+2\varepsilon=3\varepsilon.$$

$\textbf{Remark.}$ We only used that the function $f$ is bounded and continuous at $1$. The proof heavily relied on the existence of the function (sequence) $k$. Functions, for which such function $k$ exists (like $\ln$) are called super slowly varying. They arise in Probability theory. The same proof could be carried for any such function. 

The proof proposed from Prof. Gadjev relies on the following equivalences

$$\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)\equiv\int_1^n  f\left(\frac{\ln x}{\ln n }\right)dx\underbrace{=}_{x=n^y}\int_0^1  f\left(y\right)\ln(n)n^ydy$$

 and consequently $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)=\lim_{n\to\infty}\int_0^1  \frac{\ln n}{n}n^y f\left(y\right)dy.$$

 This reasoning could be made rigorous. After that, what remains is to observe that the sequence 

$ \displaystyle\frac{\ln n}{n-1}n^y$ tends to the Dirac Delta function at $1$.

Asymptotics of the solution of a differential equation

Let $f$ be twice continuously differentiable function defined on $\mathbb{R}$, such that $f(x)f''(x)=1$ for all $x\ge 0$, $f(0)=1$ and $f'(0)=0$.

Find $$\lim_{x\to\infty} \frac{f(x)}{x\sqrt{\ln(x)}}.$$

Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I.

The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.

 Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds 

$$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$

Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$

$\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence

$$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$

and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_{n\to\infty}x_n=0.$

$\textbf{2)}$ Now $1/x_n\to +\infty$  increasingly and we can use Stolz theorem as follows:

$$\lim_{n\to\infty}nx_n =\lim_{n\to\infty}\frac{n}{\frac{1}{x_n}}=\frac{(n+1)-n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}\frac{x_{n+1}x_n}{x_n-x_{n+1}}$$

Now consider the defining equation. It can be rewritten as $(1+x_{n+1})^n=1+n x_n$ hence $ x_n=x_{n+1}+S/n$ where $\displaystyle S=\sum_{k=2}^n{n\choose k}x_{n+1}^k.$

Thus 

$$\lim_{n\to\infty}\frac{x_{n+1}x_n}{x_n-x_{n+1}}=\lim_{n\to\infty}\frac{x_{n+1}\left(x_{n+1}+\frac{S}{n}\right)}{\frac{S}{n}}=\lim_{n\to\infty}\frac{nx_{n+1}^2}{S},$$

where we have used that $x_{n+1}\to 0$. Now this can be rewritten as follows

$$\lim_{n\to\infty}\frac{nx_{n+1}^2}{S}=\lim_{n\to\infty}\frac{n}{{n\choose 2}+S'},$$

 where $\displaystyle S'=\sum_{k=3}^n{n\choose k}x_{n+1}^{k-2}>0.$ This shows that the last limit is $0$.

$\textbf{3)}$  Apply the Stolz theorem in the very same way as above -

$$\lim_{n\to\infty}n^2x_n =\lim_{n\to\infty}\frac{n^2}{\frac{1}{x_n}}=\frac{(n+1)^2-n^2}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to\infty}(2n+1)\frac{x_{n+1}x_n}{x_n-x_{n+1}}=2\lim_{n\to\infty}\frac{n^2x_{n+1}^2}{S}.$$

For convenience we work with the reciprocal limit 

$$\lim_{n\to\infty}\frac{S}{n^2x_{n+1}^2}=\lim_{n\to\infty}\frac{\sum_{k=2}^n{n\choose k}x_{n+1}^k}{n^2x_{n+1}^2}=\frac{1}{2}+\lim_{n\to\infty}\sum_{k=3}^n{n\choose k}\frac{1}{n^2}x_{n+1}^{k-2}.$$

 We would be done if we prove, that the last limit is $0$.  Denote $A_n=\displaystyle \sum_{k=3}^n{n\choose k}\frac{1}{n^2}x_{n+1}^{k-2}$. Clearly $A_n>0$. Fix $\varepsilon>0$. According to $2)$, for large enough $n$ holds $\displaystyle x_{n+1}<\frac{\varepsilon}{n}.$

Thus (after completing to the binomial formula) $$A_n<\sum_{k=3}^n{n\choose k}\frac{\varepsilon^{k-2}}{n^k}=\frac{-n \varepsilon ^2+2 n \left(\frac{n+\varepsilon }{n}\right)^n-2 n \varepsilon -2 n+\varepsilon ^2}{2 n \varepsilon ^2},$$

hence $$\limsup_{n\to\infty}A_n\le \lim_{n\to\infty} \frac{-n \varepsilon ^2+2 n \left(\frac{n+\varepsilon }{n}\right)^n-2 n \varepsilon -2 n+\varepsilon ^2}{2 n \varepsilon ^2}=\frac{-\varepsilon ^2-2 \varepsilon +2 e^{\varepsilon }-2}{2 \varepsilon ^2}.$$

This is true for arbitrary $\varepsilon>0$. Letting $\varepsilon\to 0$ in the last bound we obtain (using Taylor expansion of the exponent near $0$) that  

$$\lim_{\varepsilon\to 0}\frac{-\varepsilon ^2-2 \varepsilon +2 e^{\varepsilon }-2}{2 \varepsilon ^2}=0,$$

whence  $\displaystyle\limsup_{n\to\infty}A_n=0,$ which finishes the proof.

IMO shortlist, 1998

G7. $ABC$ is a triangle with $\angle ACB = 2 \angle ABC$. $D$ is a point on the side $BC$ such that $DC = 2 BD$. $E$ is a point on the line $AD$ such that $D$ is the midpoint of $AE$. Show that $\angle ECB + 180 = 2 ∠EBC$.

 

Solution.  Put $D$ in the center. We can rewrite everything in terms of $a$ - the number corresponding to $A$ and $b$ - the number corresponding to $B$. Thus $C$ is $-2b$, E is $-a$.
The relation between the angles at $B$ and $C$ is easily reduced to the following equation:
$$\frac{(a-b)^2}{|a-b|^2}\frac{a+2b}{|a+2b|}=\frac{b^3}{|b|^3}.$$
Squaring and representing modules with conjugates, we obtain
$$\frac{(a-b)^2}{(\bar a-\bar b)^2}\frac{a+2b}{\bar a+2\bar b}=\frac{b^3}{\bar b^3}.$$
Clearing the denominator and setting everything to the left we obtain
$$a^3\bar b^3-3a\bar b^3 b^3-\bar a^3b^3+3\bar a\bar b^2 b^3=0\ (*)$$
Similarly, the equality we want to proof is equivalent to
$$\left(\frac{\frac{-b}{|-b|}}{\frac{-a-b}{|-a-b|}}\right)^2=(-1)
\left(\frac{\frac{-a+2b}{|-a+2b|}}{\frac{b}{|b|}}\right)$$
Both of the arguments of the complex numbers on both sides are between 180 and 360, so squaring is equivalent transformation here. As above we obtain
$$\frac{(a+b)^2}{(\bar a+\bar b)^2}\frac{a-2b}{\bar a-2\bar b}=\frac{b^3}{\bar b^3}.$$
Everything on the left and expand - we obtain
$$-a^3\bar b^3+3a\bar b^3 b^3+\bar a^3b^3-3\bar a\bar b^2 b^3=0.$$
But this is just the same as $(*)$.

A problem told by Zhivko Petrov

 The following integral was proposed for homework to Applied Math, by Zhivko Petrov. 

$\textbf{Problem}.$ Evaluate

$$\int_{0}^1\frac{\ln(1-x+x^2)}{x^2-x}\text{d} x.$$

Later I communicated the problem to prof. Gadjev and he proposed a neat solution (to be presented later).

Now we elaborate a solution, on an idea proposed by prof. Babev. I would like to thank the afformentioned people, as well as David Petrov for pointing me to that idea.

$\textbf{Solution}.$ Introduce $$I(y)=\int_0^1\frac{\ln(1-y(x-x^2))}{x^2-x}\text{d}x.$$

We need to find $I(1)$. Clearly $I(0)=0$. Using differentiation under the integral sign we obtain 

$$I'(y)=\int_0^1\frac{1}{1-y(x-x^2)}\text{d}x.$$

Assuming that $y\in [0,1]$, one can easily integrate the last to obtain 

$$I'(y)=\frac{4 \arcsin\left(\frac{\sqrt{y}}{2}\right)}{\sqrt{y(4-y) }}.$$

The  latter is very easy to integrate (for example making the change $y=4t^2$) in order to obtain

$$I'(y)=\left(4 \arcsin\left(\frac{\sqrt{y}}{2}\right)^2\right)'.$$

Thus 

$$I(1)=\int_0^1I'(y)\text{d}y+I(0)= 4 \arcsin\left(\frac{\sqrt{1}}{2}\right)^2-4 \arcsin\left(\frac{\sqrt{0}}{2}\right)^2+0=4 \arcsin\left(\frac{1}{2}\right)^2=\frac{\pi^2}{9}.$$

A problem proposed by prof. Gadjev

Evaluate 

$$\int_0^{\infty}\frac{\ln x}{x^2+2x+5}\, \text{d}x$$

$\textbf{Solution.}$

The indefinite integral is not expressible in elementary functions. Denote the integral by $I$. First make the change of variables $\displaystyle x\to \frac{1}{x}$ to obtain 

$$I=-\int_0^{\infty}\frac{\ln x}{5x^2+2x+1}\, \text{d}x\ \ $$

In the original integral make the change of variables $x\to 5x$ to obtain 

$$ I=\int_0^{\infty}\frac{\ln (5x)}{25x^2+10x+5}5\, \text{d}x=\int_0^{\infty}\frac{\ln x}{5x^2+2x+1}\, \text{d}x+\int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x=$$

$$-I+\int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x$$

whence 

$$I=\frac{1}{2} \int_0^{\infty}\frac{\ln 5}{5x^2+2x+1}\, \text{d}x=\frac{1}{8}\arctan(2)\ln(5)$$

Geometric-analytic problem

Let $f:\mathbb{R}^2\to\mathbb{R}$ be strictly positive Lipschitz function with constant $1/2$. Let $A$ be a nonempty subset of $\mathbb{R}^2$, such that if $x\in A$ and $y\in\mathbb{R}^2$ with $\|x-y\|=f(x)$, then $y\in A$. Prove that $A=\mathbb{R}^2$.

The problem was proposed to Bulgarian TST, 2009. Do you know earlier source or some context of the problem?

A problem proposed by prof. Babev

Consider the sequence
$$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$
It could be proven that
$$\lim_{n\to \infty}n a_n=\frac{1}{2}$$
 Now we are going to prove that
$$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$
and show how to derive all such limits.
$\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as
$$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$
$$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$
Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to
$$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$
Since for $s\in[0,1]$, $\displaystyle \frac{n}{n+s}$ is bounded between $1$ and $ \displaystyle \frac{n}{n+1}$, the last limit is bounded between the limits
$$ \lim_{n\to \infty}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s\ \ \mbox{and}\ \ \lim_{n\to \infty}\frac{n}{n+1}\int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s$$
which both are equal to $\displaystyle  \int_{0}^{1}\sin^2(\pi s)(-s)\text{d}s=-\frac{1}{4}$, hence the result follows.

$\textbf{Comment.}$ In a similar vein it could be proven that
$$\lim_{n\to\infty}n\left(n\left(na_n-\frac{1}{2}\right)+\frac{1}{4}\right)=\int_0^1\sin^2(\pi s)s^2\text{d}s$$
and we could extend this further.
This could be viewed as a "Taylor series of the sequence at infinity", namely
$$a_n=\sum_{j=1}^\infty b_j\left(\frac{1}{n}\right)^j$$
where $$b_j= \int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\text{d}s$$
Having this perspective leads to general and more insightful solution. Consider the series
$$\sum_{j=1}^\infty b^j \left(\frac{1}{n}\right)^j=\sum_{j=1}^\infty\int_{0}^{1}\sin^2(\pi s)(-s)^{j-1}\left(\frac{1}{n}\right)^j\text{d}s=\int_{0}^{1}\sin^2(\pi s)\frac{1}{n+s}\text{d}s.$$
(where we have exchanged summation and integration and then calculated the sum of the geometric progression)
After changing the variables $s\to t-n$ we arrive at the integral
$$\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t$$
which is exactly $a_n$.

Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$

Proof. The function
$$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$
is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that
$$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$
If we make change of variables in the latter integral $t\to 1-t$ we obtain
$$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$
On the other hand, using integration by parts, we obtain
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$
Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$
Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain
$$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$
On the other hand  $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t+\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=\int_{0}^1\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{6}$$
Substracting the last two equalities we obtain  $$\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{12}-\frac{1}{2}\log\left(\frac{1}{2}\right)^2$$
and thus $\displaystyle S\left(\frac{1}{2}\right)=\frac{\pi^2}{12}+\frac{1}{2}\log\left(\frac{1}{2}\right)^2$.

Some math jokes (which turn out to be correct)

Prove that $$\int_{0}^{\infty}\frac{1}{(x^2+1)(x^{\pi}+1)}\text{d}x=\int_{0}^{\infty}\frac{1}{(x^2+1)(x^{e}+1)}\text{d}x$$
Prove that $$\int_{0}^{\pi/2}\frac{1}{1+\tan^{\pi}(x)}\text{d}x=\int_{0}^{\pi/2}\frac{1}{1+\tan^{e}(x)}\text{d}x$$

$\textbf{Proof.}$ Both of the integrals are being solved, when the peculiar powers are being replaced by a parameter "a" and then differentiation with respect to "a"  is performed. Actually the integrals are related vie the change of variables $x\to\arctan(x)$.
For the first integral, denote
$$f(x,a)= \frac{1}{(x^2+1)(x^{a}+1)}.$$
Then $$\frac{\partial }{\partial a}f(x,a)=\frac{x^a \log (x)}{\left(x^a+1\right) \left(x^a+1\right)^2}$$
Consider the integral
$$\int_{0}^{1}\frac{\partial }{\partial a}f(x,a)\text{d}x.$$ Making the change $x\to 1/x$ this transforms to the integral
$$\int_{\infty}^{1}\frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{-1}{x^2}\text{d}x=\int_{1}^{\infty}\frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{1}{x^2}\text{d}x$$
Now it remains to observe that
$$ \frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{1}{x^2}=-\frac{\partial }{\partial a}f(x,a)$$
so that
$$\int_{0}^{1}\frac{\partial }{\partial a}f(x,a)\text{d}x=-\int_{1}^{\infty}\frac{\partial }{\partial a}f(x,a)\text{d}x$$
and hence
$$\int_{0}^{\infty}\frac{\partial }{\partial a}f(x,a)\text{d}x=0.$$
Thus  $$\int_{0}^{\infty}f(x,a)\text{d}x$$
is a constant independent of $a$, which proves the equality. Plugging $a=0$ we see moreover that the value of the integrals is $\displaystyle \frac{\pi}{4}$.

$\textbf{Addendum}$. Another solution proceeds as directly making the change of variable $x\to 1/x$. Thus

$$\int_0^\infty\frac{1}{(x^2+1)(x^{a}+1)}\text{d}x=\int_0^\infty\frac{x^a}{(x^2+1)(x^{a}+1)}\text{d}x.$$
Summing both integrals yields the solution.

The second problem could be done in a similar manner.