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For $f \in L^2([0,1])$, define operator $Tf: x \mapsto \frac{1}{x}\int_0^x f(y)dy$. Show that $T$ is not a compact operator on $L^2([0,1])$ and that $T$ is bounded.

For the second part, I can show $T$ is bounded by looking at $\|Tf\|_2$ and rewriting it by integration by parts and then apply the Cauchy-Schwartz inequality. However, I was not able to find a bounded sequence of $L^2$ functions so that its image under $T$ is not precompact in $L^2$.

Any help is tremendously appreciated.

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    $\begingroup$ Hint: try $f_n(x) = \sqrt{n} 1_{[0,1/n]}(x)$. A helpful fact is that if a sequence converges in $L^2$ then it has a subsequence which converges almost surely, so you can identify the possible limit points of $T f_n$. $\endgroup$ – Chris Janjigian Jul 23 '18 at 0:22
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For $f_n(x) = \sqrt n \chi_{(0, \frac 1 n]}(x)$ we have \begin{eqnarray*} Tf_n(x) & = & \frac 1 x \int_0^x\sqrt n\chi_{(0, \frac 1 n]}(y)\; {\rm d} y\\ & = & \begin{cases} \sqrt n & \text{ if }0< x \leq \frac 1 n\\ \frac{1}{x\sqrt n} & \text{ if } \frac 1 n< x< 1. \end{cases} \end{eqnarray*} From this formula we see that $\lim_{n\to\infty}Tf_n(x)\to 0$ pointwise for all $0< x< 1$. Therefore, if there were some $L^2$-convergent subsequence of $Tf_n$, say $(Tf_{n_k})_{k = 1}^\infty$, then we must have $Tf_{n_k}\to0$ in $L^2([0, 1])$. However, the following computation shows that no member of the sequence $(Tf_n)_{n = 1}^\infty$ can get anywhere near its potential subsequential limit. For every $n\geq 1$, \begin{eqnarray*} \|Tf_n\|_{L^2([0, 1])}^2 & = & \int_0^{1/n}(\sqrt n)^2\; {\rm d}x + \int_{1/n}^1\left(\frac 1{x\sqrt n}\right)^2\; {\rm d}x \\ & = & 1 + \frac 1 n\int_{1/n}^1x^{-2}\; {\rm d} x \\ & = & 2 - \frac 1 n\\ & > & 1. \end{eqnarray*}

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Another approach using the same sequence $f_n = \sqrt{n}\chi_{[0, 1/n]}$.

Here it is shown that $(f_n)_n$ converges to $0$ weakly:

  • $\|f_n\|_2 = 1, \forall n\in\mathbb{N}$ so $(f_n)_n$ is bounded

  • for $g \in L^\infty[0,1] \cap L^2[0,1]$ we have $$|\langle f_n, g\rangle| \le \| g \|_\infty \cdot \sqrt{n} \int_0^{1/n}\! dx = \frac{\| g \|_\infty}{\sqrt{n}} \to 0$$ and $L^\infty[0,1] \cap L^2[0,1]$ is dense in $L^2[0,1]$.

If $T$ were compact, it would map weakly convergent sequences to strongly convergent sequences so we would have $Tf_n \to 0$ strongly.

But $\|Tf_n\|_2 = \sqrt{2-\frac1n}$ which doesn't converge to $0$.

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