Somemathhere

In what follows, we will prove that every Hilbert-Schmidt integral operator is compact. Formally, let $k\in L_2([0,1]\times[0,1])$ and $T:L_2[0,1]\to L_2[0,1]$ be defined by $$Tf(x):=\int_0^1k(x,y)f(y)dy.$$ Then $T$ is compact operator. Although the proof may seem a bit long at first sight, it is proceeds naturally, utilizing approximation of $k$ with continuous functions. $\textbf{Proof}$. We approximate $k$ with a sequence of continuous functions, then prove that each of the corresponding operators has image contained in $C[0,1]$, then show that each of these operators is compact when $C[0,1]$ is equipped with the $\sup$ norm. This will be done via the Arzela-Ascoli theorem. Finally show that these converge uniformly (in the operator norm) to the original operator, and since they are compact, the original must be as well. The Cauchy-Bunyakovsky-Schwartz inequality is used throughout the proof. Thus we start by choosing a sequence $(k_n)_{n\ge 1}$ of continuous functions con...

Let $X$ be a normed vector space. Let $(x_n)_{n\ge 1}$ be a sequence in $X$, such that for any $f\in X^*$ $$\sum_{n=1}^{\infty}|f(x_n)|<\infty.$$ Then there exists a constant $C$ such that $$\sum_{n=1}^{\infty}|f(x_n)|\le C\|f\|, \ \forall f\in X^*$$. $\textbf{Proof}.$ Consider the operator $T:X^*\to l_1$, defined by $$T(f)=(f(x_1),f(x_2),\ldots).$$ Clearly $T$ is well-defined linear operator. Moreover $T$ is bounded. We shall check that using the Closed-graph theorem ($X^*$ is Banach, even if $X$ is not). For this purpose take a sequence $(f_k)_{k\ge 1}\subset X^*$ converging to some $f$, and let $Tf_n\rightarrow p$ for some $p\in l_1$. Our aim is to show that $Tf=p$. Since $Tf_k\rightarrow p$ (in $l_1$), we have that $$\sum_{n=1}^{\infty}|f_k(x_n)-p_n|\rightarrow 0$$ as $k$ goes to $\infty$. This means that for any $n\in\mathbb{N}$ we have $$\lim_{k\to\infty}f_k(x_n)=p_n.$$ On the other hand, as $f_k\rightarrow f$, we have that $$\lim_{k\to\infty}f_k(x_n)=f(x_n).$$ ...