Is $(Tf)(x):=\int_0^x f(t)dt$ defined on $L^2([0,1])$ a finite rank operator?

Question

Let $T:L^2([0,1])\to L^2([0,1])$ defined as $(Tf)(x):=\int_0^x f(t)dt$, is $T$ a finite rank operator?

Thank you for your help. I have the feeling it is. But How can I show it?

@CameronWilliams don't know what's the standard orthonormal basis for $L^2([0,1])$... — MorganeMaPh, Aug 09 '16 at 15:16
@WillieWong Yes, Cauchy-Schwarz, we are on a compact set. Thanks! — Olivier Oloa, Aug 09 '16 at 15:23
The image of $T$ contains all $C^1$ functions which vanish at zero. — Luiz Cordeiro, Aug 09 '16 at 15:23

score 5 · Answer 1 · answered Aug 09 '16 at 15:21

Consider the functions $$ f_n(t)=t^n \qquad (n\in\mathbb{N}). $$ Then $\{f_n:n\in\mathbb{N}\}$ is linearly independent, and we have $$ (Tf_n)(x)= \int_0^xt^n\ dt=\frac{1}{n+1}x^{n+1}.$$ Thus, $\{Tf_n:n\in\mathbb{N}\}$ is a linearly independent infinite set of vectors in $L^2([0,1])$, hence $T$ is not finite-rank.

score 1 · Accepted Answer · answered Aug 09 '16 at 15:37

While it's pretty clear that $T$ does not have finite rank it's easy to see that $T$ is compact (hence the norm limit of a sequence of operators of finite rank).

For example, say $f_n$ is bounded. Passing to a subsequence we can assume that $f_n\to f$ weakly. Hence $Tf_n\to Tf$ pointwise. But $(Tf_n)$ is also equicontinuous. So $Tf_n\to Tf$ uniformly, hence $Tf_n\to Tf$ in norm.

Is $(Tf)(x):=\int_0^x f(t)dt$ defined on $L^2([0,1])$ a finite rank operator?

2 Answers2