Show that $T: L^2([0, \infty )) \to C_0([0,\infty))$ is compact operator. Let $C_0 \left( [0,\infty) \right)$ be the set of continuous functions on $[0,\infty)$ vanishing at $\infty$, which is Banach space with $\lVert f \rVert_\infty = \underset{t \in [0,\infty)}{\sup} |f(t)|$. Define $T \in B\left(L^2([0,\infty)), C_0 ([0,\infty)) \right)$ by
$$Tf(t) = \frac{1}{1+t} \int_0^t f(s) ds.$$
Show that $T$ is a compact operator.
Hint: To show that $TB_{L^2 [0,\infty)}$ is totally bounded, take sufficiently large $R >0$ such that $\underset{t \ge R}{\sup} |Tf(t)|$ is uniformly small for $f \in B_{L^2[0,\infty)}$, and then apply the Ascoli-Arzela theorem to the restriction of $TB_{L^2[0,\infty)}$ to $[0,R]$. I do not know how to apply this hint.
Edited: I have seen some similar questions in the web, but all of them was on closed intervals. In this case the interval is not bounded, so at first -as I know- I have to prove that $T$ is bounded, which is not realy clear when the interval is not bounded. And After that I need to prove the compactness of the operator, by definition or by theorems, Could you please show me steps (hints) that i can follow.
My proof till now:
First let's show the boundedness of $T$.
$|Tf(t)| \le \frac{1}{1+t} \int_0^t f(s)ds \le \|f\|_2 \frac{t}{1+t}$. So, $\|T\| \le 1$
Is is OK?
 A: (Never mind the crossed-out text; based on the hint, you have not learned what it assumes) The easiest approach is to use the "completely continuous" characterization of compactness: if $X$ is reflexive, then $T\in L(X\to Y)$ is compact iff $f_n\rightharpoonup0\Rightarrow Tf_n\to0$.  $L^2$ is reflexive, so you want to show that, if $$\int_0^{\infty}{f_n(x)g(x)}\to0\text{ for all }g\in L^2$$ then $Tf_n\overset{C_0}{\to}0$.  Now expand the definition of $T$ and choose $g$ cleverly.
If you haven't seen that characterization, then you'll need a tool to characterize compact subsets of $C_0$.  In the problems you've seen, the interval was closed (and thus compact) so you could apply Arzelà-Ascoli.  Of course, $C_0$ is not defined on a compact interval. It turns out that for noncompact spaces, one can obtain an analogous result, as long as one has uniform decay at each end.  So here's a simple sufficient condition:Fn. 1

Suppose that

*

*$f_n\to0$ uniformly on any compact set $K\subseteq[0,\infty)$ and

*there is a constant $A$ independent of $n$ such that $|f_n(x)|\leq\frac{A}{\sqrt{x}}$.

Then $\{f_n\}_n$ has a subsequence converging in $C_0$ to $0$.

Using a strengtheningFn. 2 of the boundedness argument you gave in comments, this shows that $T$ sends the unit ball to a sequentially-compact set.
Fn. 1 This is not a standard result, so if this is a homework question, make sure to prove this as a lemma in your solution.  This is where you apply the hint.
Fn. 2 More than just fixing the flaw Andrew McMillan pointed out.
A: You will first need the following which follows from Cauchy Schwartz:
$$\left|\int_x^yf(s)ds\right|=\left|\int_{[x,y]}f(s)1ds\right| \leq ||f||_2||x-y||^{\frac{1}{2}}. $$
Now you have: $$\left|Tf(t)\right|\leq \frac{\sqrt{t}}{1+t}||f||_2 $$
As $\left|\frac{\sqrt{t}}{1+t}\right|\to 0$ as $t\to \infty $ and fixing $||f||_2\leq 1$ , given $\epsilon>0$, we can find $R$ so that for all $t>R$, $\left|Tf(t)\right| <  \epsilon$
Now for $f\in L_2([0,\infty)), ||f||_2\leq 1 $ we have that $Tf$ is bounded. Further, direct application of our first inequality shows that the following is equicontinuous:
$$\{ Tf:[0,R]\to \mathbb{R}: ||f||_2\leq 1\} $$.
So Arzelà–Ascoli theorem applies and the above is actually precompact and hence totally bounded. However, we need, $\{ Tf:[0,\infty)\to \mathbb{R}: ||f||_2\leq 1\} $ to be totally bounded. Fortunately, we can extend any finite $\epsilon$ cover of the former to  a $2\epsilon$ cover of the the latter by noticing for $t>R$ and $||f||_2,||g||_2\leq 1$
$$ \sup_{t>R}\{|Tf(t)-Tg(t)|\}<2\epsilon$$ hence $\{ Tf:[0,\infty)\to \mathbb{R}: ||f||_2\leq 1\} $ is totally bounded and hence $T$ is compact.
