Show that $f \overset{T}{\rightarrow} \frac{1}{x} \int_0^x f(t) dt$ is Bounded, and is NOT Compact in $L^2(0, \infty)$

Question

Can someone help me with this question?

Let $f \in X = L^2(0, \infty)$, and define \begin{equation} (Tf)(x) = \frac{1}{x} \int_0^x f(t) dt \ . \end{equation} Show that $T$ is Bounded, and is NOT Compact.

Thanks in advance.

Edit: For boundedness, I tried: \begin{eqnarray} \| f \| \leq 1 \Rightarrow \| Tf \|^2 &=& \int_0^{\infty} \lvert (Tf)(x) \rvert ^2 \ dx \\ &=& \int_0^{\infty} \lvert \frac{1}{x} \int_0^x f(t) \ dt \rvert ^2 \ dx \\ &\leq& \int_0^{\infty} \frac{1}{x^2} \int_0^x \lvert f(t) \rvert ^2 \ dt \ dx \\ &\leq& \int_0^{\infty} \frac{1}{x^2} \| f \| ^2 \ dx \\ &=& \| f \| ^2 \int_0^{\infty} \frac{1}{x^2} \ dx \\ \end{eqnarray} But the problem is that $\int_0^{\infty} \frac{1}{x^2} \ dx$ is NOT finite.

To prove that $T$ is not compact, I think we should give a sequence of functions, $\{ f_n \}_{n=1}^{\infty}$, such that for every $n$, we have $\| f_n \| \leq 1$ and $\{ Tf_n \}$ has no Cauchy subsequence. But I can't find such sequence of functions.

Answer 1

I thought I would just offer an alternate solution using Minkowski's integral inequality applied to:

$$ \| Tf \|_2 = \left\{ \int_0^{\infty} \left( \int_0^x x^{-1} f(t)\,dt \right)^{2}\,dx \right\}^{1/2} $$

But first we do a variable substitution $t \rightarrow xt$ so the inner integral is integrating over a fixed space:

$$ \| Tf \|_2 = \left\{ \int_0^{\infty} \left( \int_0^1 f(xt)\,dt \right)^{2}\,dx \right\}^{1/2} $$

Minkowski's inequality then yields:

$$ \| Tf \|_2 \leq \int_0^1 \left \{ \int_0^{\infty} f^2(xt)\, dx \right\}^{1/2}\, dt $$

Now undo our variable substitution, this time sending $x \rightarrow x/t$, and get:

$$ \| Tf \|_2 \leq \int_0^1 \left \{ \int_0^{\infty} t^{-1} f^2(x)\, dx \right\}^{1/2}\, dt = \left( \int_0^1 t^{-1/2}\,dt \right)\left ( \int_0^{\infty} f^2(x)\,dx\right )^{1/2}$$

And, of course, $\int_0^1 t^{-1/2}\,dt$ is finite.

To answer your second question. Let

$$ f(x) = \begin{cases} & 0 & x \leq 0 \\ & 1 & 0 < x < 1/2 \\ & -1 & 1/2 \leq x \leq 1 \\ & 0 & 1 \leq x \end{cases} $$

So $\|f\|_2 = 1$. Now let $f_n(x) = n^{-1/2}f\left(\frac{x - n}{n}\right)$. So,

$$\|f_n\|_2 = n^{-1/2} \left( \int_n^{2n} dt \right)^{1/2} = 1.$$

The important property here is that $\int_0^1 f(t) dt = 0$. This is going to ensure that $Tf_n$ have disjoint support. Why? Recall:

$$ Tf_n(x) = \frac{1}{x}\int_0^x n^{-1/2} f\left(\frac{t-n}{n}\right)\,dt $$

Since $f(x)$ is $0$ for $x < 0$ we have that $Tf_n$ is 0 for $x < n$. And since $\int_0^1 f(t)\,dt =0 $, we have:

$$ Tf_n(2n) = \frac{1}{2n} \int_n^{2n} f\left(\frac{t - n}{n}\right)\,dt = \frac{1}{2n} \int_0^n f\left(\frac{t}{n}\right)\,dt = 0$$

And you can see that $Tf_n(x) = 0$ for $x > 2n$. So if we take $f_{2^k}$ as $k = 1, 2, 3, \dots$ we can see that $Tf_{2^k}$ have disjoint supports. So to prove that these have no convergent subsequence you just need to show that $\|Tf_k\|_2$ is bounded below for any integer $k$. We will estimate $\| Tf_k \|_2^2$ but only focus on those x where $f_k$ is positive.

$$\| Tf_k \|_2^2 = \int_k^{2k} \frac{1}{x^2} \left\{ \int_0^x f_k(t)\,dt \right\}^2\,dt \geq \int_k^{3k/2} \frac{1}{4k^2} \left\{\int_k^x f_k(t)\,dt \right\}^2\,dx $$

We have:

$$ \int_0^x f_k(t)\,dt = k^{-1/2}(x - k) $$

for $k \leq x \leq 3k/2$. So we get:

$$\| Tf_k \|_2^2 \geq \frac{1}{4k^2} \frac{1}{k} \int_k^{3k/2} (x - k)^2\, dx$$

And this can be shown to be bounded below be something like $1/100$. Sorry, but nothing else comes to mind for proving this.

Answer 2

For the boundedness, observe that

\begin{align*} \| Tf \|_2^2 &= \int_{0}^{\infty} |Tf(x)|^2 \, dx \\ &\leq \int_{0}^{\infty} \frac{1}{x^2} \int_{0}^{x} \int_{0}^{x} |f(u)||f(v)| \, dudvdx \\ &= \int_{0}^{\infty}\int_{0}^{\infty} |f(u)||f(v)| \left( \int_{u \vee v}^{\infty} \frac{dx}{x^2} \right) \, dudv \\ &= \int_{0}^{\infty}\int_{0}^{\infty} \frac{|f(u)||f(v)|}{u \vee v} \, dudv \\ &= \int_{0}^{\pi/2}\int_{0}^{\infty} \frac{|f(r\cos\theta)||f(r\sin\theta)|}{\cos\theta \vee \sin\theta} \, drd\theta \\ &\leq \int_{0}^{\pi/2} \frac{ \| f(r\cos\theta) \|_{L^2(dr)} \| f(r\sin\theta) \|_{L^2(dr)} }{\cos\theta \vee \sin\theta} \, d\theta \\ &= \int_{0}^{\pi/2} \frac{\| f\|_2^2}{(\cos \theta \vee \sin\theta) \sqrt{\cos\theta\sin\theta}} \, d\theta, \end{align*}

where $a \vee b := \max \{a, b \}$ is the shorthand notation for the maximum function. Then it is not hard to see that the last integral is bounded.

Answer 3

To show noncompactness, let $a> 0$ and set $f_a(x)= (a/2)^{-1/2}\chi_{(a/2,a)}.$ Then $f$ is a unit vector in $L^2.$ It is straightforward to see

$$T(f_a)(x) = \begin{cases} 0,\, 0 < x < a/2 \\ (a/2)^{1/2}/x,\, x> a.\end{cases}.$$

Suppose $a<b/2.$ Because $T(f_b) = 0$ on $(0,b/2),$

$$T(f_a)(x)-T(f_b)(x) = (a/2)^{1/2}/x, a<x<b/2.$$

Thus

$$\int_0^\infty|T(f_a)(x)-T(f_b)(x)|^2\, dx \ge \int_a^{b/2}|T(f_a)(x)|^2\, dx =\frac{1}{2} -\frac{a}{b}.$$

This is $\ge 1/4$ if $a/b\le 1/4.$ It's now easy: The unit vectors $f_{1/4^n}$ satisfy

$$\|T(f_{1/4^n}) - T(f_{1/4^m})\|_2 \ge \frac{1}{2}$$

if $m\ne n.$ Thus no subsequence of $T(f_{1/4^n})$ can be Cauchy. Hence $T$ is not compact.