# Why the Cesàro operator $(Cf)(s) = \frac{1}{s}\int_0^sf(t)dt$ on $L^2(0,\infty)$ is not compact?

By Hardy's inequality, $$C$$ is bounded. But I find no way to prove it to be noncompact. I was thinking about its adjoint $$(C^\ast f)(s) = \int_x^\infty\frac{1}{t}f(t)dt$$

But I'm afraid it is equally difficult to show that $$C^\ast$$ is noncompact. So this approach does not work.

I also attempted to use a classic sequence on the unit ball: $$f_n = \chi_{[n,n+1]}$$. But it turned out to be a monster. Indeed we have $$\lVert f_n \rVert = 1$$ and $$\lVert f_m-f_n \rVert = \sqrt{2}\delta_{mn}$$, which is pretty neat. But computing the norm of $$Cf_n$$ and $$C(f_m-f_n)$$ is a disaster: $$\lVert Cf_n \rVert = \sqrt{2-2n\ln\left( 1 + \frac{1}{n}\right)} \\ \lVert C(f_m-f_n) \rVert =\sqrt{\frac{2n+1}{n+1}-2n\ln\left( 1+\frac{1}{n}\right)+2+\frac{1}{m}-2(m+1)\ln\left( 1+\frac{1}{m}\right)+\ln\frac{m}{n+1}}$$ provided $$m>n$$ (hopefully my calculation wasn't wrong!). With these being said, is there any accessible way to prove the noncompactness (whether direct or indirect)? I guess I found two inaccessible way (or maybe not as I didn't realise how to use it). Many thanks in advance!

Here is one way to show that the Hardy transform $$H:f\mapsto\frac1t\int^t_0f$$ as an operator on $$L_p(0,\infty)$$, where $$1, is not compact. Consider the family of functions $$\{\phi_A(x)=A^{1/p}\mathbf{1}_{(0,1/A]}(x): A>0\}$$. We have that $$\|\phi_A\|_p=1$$ and $$G_A(x):=(H\phi_A)(x)=\frac1x\int^x_0\phi_A = A^{1/p}\Big(\mathbf{1}_{(0,1/A]}(x) + \frac{1}{Ax}\mathbf{1}_{(1/A,\infty)}(x)\Big)$$ Then, for $$A, \begin{align} |G_A - G_B|&=\big(B^{1/p}-A^{1/p}\big)\mathbb{1}_{(0,1/B]}(x)+\left|\frac{B^{1-\tfrac1p}}{x}- A^{1/p}\right|\mathbb{1}_{(1/B,1/A]}(x) \\ &\qquad + \big(A^{\tfrac1p -1} -B^{\tfrac1p -1}\big)\frac{1}{x}\mathbb{1}_{(1/A,\infty)}(x)\\ &\geq \big(A^{\tfrac1p -1} -B^{\tfrac1p -1}\big)\frac{1}{x}\mathbf{1}_{(1/A,\infty)}(x) \end{align} whence we conclude that $$\|G_A - G_B\|_p \geq \frac{1}{(p-1)^{1/p}}\left|1-(A/B)^{1-\frac1p}\right|$$ It follows that $$\{g_n=\phi_{2^n}:n\in\mathbb{N}\}$$ is sequence in unit ball in $$L_p(0,\infty)$$ for which $$\{Hg_n:n\in\mathbb{N}\}$$ has no convergent subsequence in $$L_p(0,\infty)$$.

• Thank you for your help, but I'm afraid maybe there is something wrong with this reply... For example your definition of $G_A$ is wrong as the integration should be $(0,x]$ instead of $(0,\infty)$. Do consider review and correct the answer so that it's helpful for not only me but also many others!
– user614535
Jun 27, 2021 at 6:50
• I have another question: Is it legit here to consider $1/(1-p)^{1/p}$? Since $p>1$ we would get an imaginary number...
– user614535
Jun 27, 2021 at 6:59
• Sorry to take your time but I still have some trouble. If we put $A=4$, $B=9$, $p=2$ and $x=1/6$, then $G_A(x)=G_B(x)=2$. But $\frac{1}{B} < x < \frac{1}{A}$, hence $|G_A-G_B|< |\frac{1}{2}-\frac{1}{3}| \cdot 6 = 1$. The crutial inequality is not correct. (If I'm wrong, just correct me!)
– user614535
Jun 27, 2021 at 7:36
• If I'm right we have $$G_A-G_B=\left(\sqrt{B}-\sqrt{A} \right)\chi_{(0,1/B]}(x)+\left(\frac{1}{\sqrt{B}x}-\sqrt{A}\right)\chi_{(1/B,1/A]}(x) +\left(\frac{1}{\sqrt{B}} - \frac{1}{\sqrt{A}} \right)\frac{1}{x}\chi_{(1/A,+\infty)}(x)$$ Hence we can only get $|G_A-G_B| \geq \left|\frac{1}{\sqrt{B}} - \frac{1}{\sqrt{A}} \right|\frac{1}{x}\chi_{(1/A,+\infty)}(x)$. An important thing is that $\frac{1}{A}>\frac{1}{B}$.
– user614535
Jun 27, 2021 at 7:41
• @ZoeDesvl: I added more details in the derivation of the inequality concerning $|G_A-G_B|$ ($0<A<B$). You were correct and I made s silly mistake before. I hope this is it! Jun 27, 2021 at 16:14