# $T: L^2[0,1] \to L^2[0,1]$, $Tf(x)= \frac{1}{x}\int_{0}^x f(y)$, is a bounded but not compact operator.

To show that the image of $$T$$ lies in $$L^2$$, and derive its bound, I tried the following: $$\|Tf\|_{2} = \left(\int|\int \frac{1}{x}f(y)dy|^2dx\right)^{\frac{1}{2}} \leq \int \sqrt{\int|\frac{1}{x}f(y)1_{[0,x]}(y)|^2dx}dy = \int_{0}^{1} \sqrt{ \frac{1}{y}-1 }|f(y)|.$$ Then probably using Holder's inequality, but this does not seem to work. Then to show it is not a compact operator, I want to construct a bounded sequence $$f_n$$ where $$Tf_n$$ does not have a convergent subsequence.

• I edited the title, because I'm all but certain there was an error. By the way, how do you justify bringing the square root inside the integral? Commented Sep 16, 2018 at 15:33
• This is the Hardy operator: math.stackexchange.com/q/587173/144766 Commented Sep 16, 2018 at 15:46
• I am using the integral version of the Minkowski inequality. @Aweygan Commented Sep 16, 2018 at 16:31

For the continuity of $T$, you can also check the link by mechanandroid.

For compactness, however, this link sends you to this question, where it is shown that $T : \mathcal{C} ([0,1]) \to \mathcal{C} ([0,1])$ is not compact, but $T : \mathcal{C} ([0,1]) \to \mathbb{L}^2 ([0,1])$ is compact. Of course, the functions used there are not suitable to show the non-compactness of $T : \mathbb{L}^2 ([0,1]) \to \mathbb{L}^2 ([0,1])$. Hence this question does not seem to be a duplicate.

Now, what would a suitable sequence of function $(f_n)$ look like? We want $\|f_n\|_{\mathbb{L}^2} \equiv 1$, an $T(f_n)$ as large as possible (so as to avoid convergence to $0$). The first thing is to avoid cancellations in the integral, since it makes $T(f_n)$ smaller. So let us look for non-negative $f_n$.

Then, we would like to put the most possible mass close to $0$; then $\int_0^x f_n (t) dt$ will be quite large for a small value of $x$, which makes $T(f_n)$ large. So, a good try is to take $f_n (t) := \sqrt{n} \mathbb{1}_{[0,1/n]} (t)$, which has unit norm. Then:

$$T(f_n) (t) = \left \{ \begin{array}{ccc} \sqrt{n} & \text{if} & t \in [0,1/n] \\ 1/(\sqrt{n}t) & \text{if} & t \in [1/n,1] \end{array}\right. .$$

We compute $\|T(f_n)\|_{\mathbb{L}^2}^2 = 1+\int_{1/n}^1 1/(nt^2)dt = 2-1/n$.

In addition, $(T(f_n))_{n \geq 0}$ converges almost everywhere to $0$, so any limit point of this sequence must be $0$. Since the norm of $T(f_n)$ converges to $2$, this cannot happen, so $(T(f_n))_{n \geq 0}$ has no limit point.

• Interesting solution for non-compactness, but I have got one question: how did you understand that you should find the sequence with such property? (I.e I’m wondering the initial motivation for these idea) Commented Mar 1, 2020 at 12:44
• And what construction should be used for case when $p=1$? Commented Mar 2, 2020 at 16:31

Let us prove that $T$ is a bounded operator, and $\|T\| \leq 2$.

By density, it is enough to show that $$(1) \qquad \|Tf\|_2 \leq 2 \|f\|_2 \qquad \forall f \in C^\infty_c((0,1)).$$ Let $F(x) := \int_0^x f(y)\, dy$. Clearly, $F\in C^1$ and $F' = f$. Integrating by parts and using Cauchy-Schwarz, we have that $$\begin{split} \|Tf\|_2^2 & = \int_0^1\frac{1}{x^2} F(x)^2\, dx = - F(1)^2 + 2 \int_0^1 f(x) \frac{1}{x}\, F(x)\, dx \\ &\leq 2 \int_0^1 f(x) \ Tf(x)\, dx \leq 2 \|f\|_2 \|Tf\|_2, \end{split}$$ so that (1) follows.

• Can you explain more on the by "density" part? Commented Sep 16, 2018 at 16:33
• Let $f\in L^2$, and let $(f_n)\subset C^\infty_c$ be a sequence converging to $f$ in $L^2$ and pointwise (a.e.). Since $(Tf_n)$ is Cauchy in $L^2$ (by (1)) and converges pointwise to $Tf$, we get that $Tf_n \to Tf$ in $L^2$. This should be enough to conclude. Commented Sep 16, 2018 at 16:48