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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, $A=0$.

$\textbf{Problem}.$ 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 ...

$\textbf{Theorem.}$ 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}(\{...

 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 $\hat X$.

 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 $$\sum_{n=1}^\infty a_nb_n$$ i...

$\textbf{Problem 1.}$ 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 $$f(x+y)\ge 2022^xf(x)+f(y).$$