# Show that $\int_0^x f(t)dt$ is compact and injective

Exercise:

Consider $T:L^2([0,1])\to L^2([0,1])$ given by $\int_0^x f(t)dt$ with $x\in[0,1]$. Show that $T$ is compact and infective.

Is my solution correct? My solution:

By Arzéla-Ascoli it suffices to show that $T$ is closed, bounded and equicontinuous.

• Boundedness: $\Vert Tf \Vert\leq (\int_0^x 1 dt)^{1/2}(\int_0^x f^2(t)dt)^{1/2}\leq \Vert f\Vert_2$ so that $\Vert T\Vert\leq 1$.
• T is closed: let $(f_n,Tf_n)$ be a sequence in the Graph of T s.t. $f_n\to f$ and $Tf_n\to g$. So we have to show that $g=Tf$. Thus $\forall \varepsilon$ $\exists N$ such that $\Vert Tf_n-g\Vert_2\leq \varepsilon$ for all $n>N$. \begin{equation} \Vert Tf_n-g\Vert_2^2\leq=\int_0^1\left(\int_0^xf_n(t)dt-g(x)\right)^2dx=\int_0^1\left(\int_0^x\vert f_n(t)-g'(t)\vert dt\right)^2dx \end{equation} and so by uniqueness of the limit $g'=f$. Thus $Tf=Tg'=g$.
• Equicontinuity: Fix $\varepsilon>0$ then it exists $\delta =\varepsilon$ s.t. if $f,g\in L^2([0,1])$ are s.t. $\Vert f-g\Vert_2\leq \delta$ then $\Vert Tf-Tg\Vert_2\leq \Vert T\Vert\Vert f-g\Vert_2\leq \varepsilon$

We see that $T$ is injective, since if $Tf(x)=0$ then it implies that $f(t)=0$ $\forall t\in [0,1]\backslash x$. But since $f$ is continuous $f(x)=0$ and thus $f\equiv 0$.

• This is not correct usage of Arzela-Ascoli. The AA theorem tells you exactly when a subset of the continuous functions is compact in the uniform norm, not the $L^2$ norm. Secondly, why would $f$ be continuous? Not every $L^2$ function is continuous. – Shalop Aug 7 '16 at 8:43
• Ok, actually in the exercise, it is not precise with which norm we must work. But the hint is given: use AA. – MorganeMaPh Aug 7 '16 at 8:44
• If you're working on $L^2$ then you're obviously using the $L^2$ norm. I will say that Arzela Ascoli can indeed be used to solve this problem correctly. First, by noting that $T$ maps $L^2$ into $C[0,1]$, and then by noting that $|Tf(x)-Tf(y)| \leq \| f \cdot 1_{[x,y]} \|_1 \leq \|f\|_2 \cdot \|1_{[x,y]}\|_2 = \|f\|_2 \cdot |x-y|^{1/2}$. This shows that $T(B)$ is equicontinuous for any ball $B \subset L^2$. Then use the fact that the $L^2$ norm is weaker than the uniform norm on $[0,1]$. – Shalop Aug 7 '16 at 8:47
• All right! Thank you @shalop , and what about the closeness? – MorganeMaPh Aug 7 '16 at 8:49
• @Shalop does your last sentence mean: $\Vert \Vert_2\leq \Vert \Vert$ with the later being the uniform norm? – MorganeMaPh Aug 7 '16 at 8:55

Let's first prove that $T$ is injective. We need to show that $Tf = 0 \implies f=0$ almost surely. Note that if $\int_0^x f=0$ for all $x$, then $f$ must be almost everywhere zero (because any finite Borel measure on $[0,1]$ is determined by its values on sets of the form $[0,x)$ with $x>0$; in this case our Borel measure is $A \mapsto \int_A f$).
Secondly, let's show that $T$ is compact. This means that if $\{f_n\}$ is a sequence in $L^2$ with $\|f_n\| \leq 1$ for all $n$, then it must be true that $Tf_n$ converges along a subsequence (w.r.t. the $L^2$ norm).
So let $f_n$ be such a sequence in $L^2$. Then we note that for all $0 \leq x<y \leq 1$, Cauchy Schwarz gives $$|Tf_n(y)-Tf_n(x)| = \bigg| \int f_n \cdot 1_{[x,y]} \bigg| \leq \|f_n \cdot 1_{[x,y]} \|_1 \leq \|f_n\|_2 \cdot \|1_{[x,y]}\|_2 \leq 1 \cdot |x-y|^{1/2}$$ This proves that $\{Tf_n\}$ is equi-Hölder(1/2), and hence is equicontinuous and pointwise bounded. By the Arzela-Ascoli Theorem ,we can conclude that the $\{Tf_n\}$ converge uniformly along a subsequence $\{Tf_{n_k}\}$. Since the uniform norm dominates the $L^2$ norm on $[0,1]$, it follows that the same subsequence also converges in $L^2$. This proves that $T$ is compact.
• also do you think we can show that $T$ is compact by showing that it is a finit-rank operator? Does it? – MorganeMaPh Aug 9 '16 at 15:06