Extend an integral inequality from $C[0,1]$ to the whole $L^{2}([0,1])$ [related to operator norm].

Question

Consider the following integral defined for $f\in L^{2}[0,1]$ and $x\in [0,1]$, that $$(T^{n}f)(x)=\dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}f(r)dr.$$

This shows that for $f\in C[0,1]$ and for any $x\in [0,1]$, we have $$|(T^{n}f)(x)|\leq \dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}|f(r)|dr\leq \dfrac{\|f\|_{\infty}}{n!}.$$ By definition of operator norm, this implies that the operator norm of $T^{n}$ $$\|T^{n}\|\leq \dfrac{1}{n!},$$ if the operator is on $C[0,1]$.

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However, I want to prove the same bound of this operator norm on the whole $L^{2}([0,1])$. Is there anyway for me to extend this proof to all of $f\in L^{2}([0,1])$? and to conclude $\|T^{n}\|\leq\frac{1}{n!}$

My attempt was to use density of continuous function of compact support in $L^{2}$ to find $g$ such that $f=g+\epsilon$ where $\epsilon>0$ is arbitrarily fixed. Then we bound the integral in the same way, and you will have a summation, the second term in the summation will go to $0$ when $\epsilon\searrow 0$. Therefore, we again have $$|(T^{n}f)(x)|\dfrac{\|g\|_{\infty}}{n!}.$$ But this cannot say anything to the operator norm $\|T\|$.

What should I do? Thank you!

Edit:

The overall purpose is to derive that $\|T\|\leq\frac{1}{n!}$. So if there is any other way to find this, it will also be really good.

Edit2: Proof

Okay, I confused myself in the first place. The operator norm has nothing to do with the sup-norm since $T:L^{2}\longrightarrow L^{2}$, so what we should do is to use Cauchy-Schwarz and $\|f\|_{L^{2}}$. See below my own answer to my post for details.

Also, Ruy gave a general theorem about how to compute the operator norm for Hilbert-Schmidt operator, and my computation will show you the idea about the proof. Basically, the absolute value of operator will be bounded by $\|f\|_{L^{2}}$ multiplication of the $L^{2}-$integral of the kernel, so we should expect such a theorem.

Thank you so much for all the users who helped me!!

Answer 1

Given $k\in L^2([0, 1]\times[0, 1])$, one may define an operator $A_k$ on $L^2([0, 1])$ by $$ A_k(f)|_x = \int_0^1 k(x, y)f(y)\, dy, \quad \forall f\in L^2([0, 1]), \quad \forall x\in [0,1]. $$

This is called an integral operator and $k$ is said to be its integral kernel.

Besides being bounded, $A_k$ is Hilbert-Schmidt, and its Hilbert-Schmidt norm $\|A_k\|_2$ may be computed as $$ \|A_k\|_2 = \left(\int_0^1 \int_0^1 |k(x, y)|^2 \, dx\, dy\right)^{1/2}. $$ See [1,Theorem VI.23] for details.

Since the operator norm is always bounded by the Hilbert-Schmidt norm, one may use the above to estimate the former.

The operator $T^n$ mentioned by the OP is clearly an example of the above, where $$ k(x, r)=\frac{1}{(n-1)!}(x-r)^{n-1}[r\leq x], $$ where $[r\leq x]$ takes the values 1 or 0 according to whether $r\leq x$ or not.

If my calculations are correct this method provides an even better estimate for the norm of $T^n$.

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[1] Reed, Michael; Simon, Barry, Methods of modern mathematical physics. I: Functional analysis. Rev. and enl. ed, New York etc.: Academic Press, A Subsidiary of Harcourt Brace Jovanovich, Publishers, XV, 400 p. $ 24.00 (1980). ZBL0459.46001.

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EDIT: OK, prompted by Calvin Khor's challenge, here is the full calculation based on the Hilbert-Schmidt norm and leading to a slightly better estimate.

I will temporarily ignore the coefficient $\frac{1}{(n-1)!}$ since it doesn't play much of a role in the computation, so let us first deal with $$ h(x, r)=(x-r)^{n-1}[r\leq x]. $$ We then have $$ \|A_h\|_2^2 = \int_0^1 \int_0^1 |h(x, r)|^2 \, dr\, dx = \int_0^1 \int_0^1 |(x-r)^{n-1}[r\leq x]|^2 \, dr\, dx = $$$$ = \int_0^1 \int_0^x (x-r)^{2n-2} \, dr\, dx = \int_0^1 \frac{-(x-r)^{2n-1}}{2n-1} \Big|_0^x\, dx = $$$$ = \int_0^1 \frac{x^{2n-1}}{2n-1}\, dx = \frac{x^{2n}}{2n(2n-1)}\Big|_0^1 = \frac{1}{2n(2n-1)}. $$ This implies that $$ \|A_k\| \leq \|A_k\|_2 = \frac{1}{(n-1)!} \|A_h\|_2 = \frac{1}{(n-1)!\sqrt{2n(2n-1)}}. $$ So we indeed get a better estimate than just $\frac{1}{n!}$ because $$ \sqrt{2n(2n-1)} \geq \sqrt{(2n-1)^2} = 2n-1 \geq n. $$

Remark. Since the Hilbert-Schmidt norm is usually much bigger than the operator norm, there is still a lot of room for improvement. Any bidders? :-)

Answer 2

Just promoting my comments to an answer because it gives the constant that appears in the question; Ruy gives a (better) method that works for more general kernels, but the kernel here is nice (it is continuous, bounded, $L^1$, ...) so we can apply generalised Minkowski as follows:

\begin{align}\left\| \int_{0}^x K(r) f(x-r)dr\right\|_{L^p([0,1])} &\le \int_0^1 \|K(r)f(x-r)\mathbf 1_{r<x}\|_{L^p (x\in[0,1])}dr \\ &= \int_0^1|K(r)| \|f(x-r)\mathbf1_{r<x}\|_{L^p (x\in[0,1])}dr \\ &\le \|K\|_{L^1([0,1])}\|f\|_{L^p([0,1])}. \end{align} So the operator norm is at most $$\|K\|_{L^1(0,1)}=\frac{\int_0^1r^{n-1}dr}{(n-1)!}=\frac1{n!}.$$

Answer 3

Okay, I guess in my way we will not be able to finally get $\frac{1}{n!}$, but a little bit worse bound $\frac{1}{\sqrt{n}(n-1)!}$, but it does not affect the general decaying I need in the end, so it is okay.

The following is my proof:

The $n-$fold iterate of $T$ is given by the formula $$(T^{n}f)(x)=\dfrac{1}{(n-1)!}\int_{0}^{x}(x-r)^{n-1}f(r)dr.$$ This implies that, for $f\in L^{2}([0,1])$, by Cauchy-Schwarz, we have the following: \begin{align*} \Big|(T^{n}f)(x)\Big|^{2}\leq \Bigg(\dfrac{1}{(n-1)!}\int_{0}^{x}|x-r|^{n-1}|f(r)|dr\Bigg)^{2}&=\Bigg(\dfrac{1}{(n-1)!}\Bigg)^{2}\Bigg(\int_{0}^{x}|x-r|^{n-1}|f(r)|dr\Bigg)^{2}\\ &\leq\Bigg(\dfrac{1}{(n-1)!}\Bigg)^{2}\|f\|_{L^{2}([0,1])}^{2}\int_{0}^{x}|x-r|^{2n-2}dr, \end{align*} and $$\int_{0}^{x}|x-r|^{2n-2}dr=\int_{0}^{x}(x-r)^{2n-2}dr=-\dfrac{1}{2n-1}\Big[(x-r)^{2n-1}\Big]_{0}^{x}=\dfrac{x^{2n-1}}{2n-1}.$$ As $x\in [0,1]$ and $n\geq 2$, we further have $$\int_{0}^{x}|x-r|^{2n-2}dr=\dfrac{x^{2n-1}}{2n-1}\leq \dfrac{1}{2n-1}\leq \dfrac{1}{n}.$$

Hence, $$|(T^{n}f)(x)|\leq \dfrac{\|f\|_{L^{2}([0,1])}}{\sqrt{n}(n-1)!},$$ and by definition of the operator norm of our operator $T:L^{2}([0,1])\longrightarrow L^{2}([0,1])$, this implies that $$\|T^{n}\|\leq \dfrac{1}{\sqrt{n}(n-1)!}.$$

Answer 4

If you are satisfied with a weaker bound $\frac1{\sqrt{n}(n-1)!}$ then we can use Cauchy-Schwarz on the $\|T^nf\|_\infty$ estimate:

\begin{align} \|T^nf\|_2 &\le \|T^nf\|_\infty \\ &\le \frac1{(n-1)!} \sup_{x\in[0,1]}\int_0^1 |x-r|^{n-1}|f(r)|\,dr\\ &\le \frac1{(n-1)!} \sqrt{\sup_{x\in[0,1]}\int_0^1 |x-r|^{2n-1}\,dr} \sqrt{\int_0^1 |f(r)|^2\,dr}\\ &= \frac1{(n-1)!} \sqrt{\frac1{2n-1}}\|f\|_2\\ &\le \frac1{\sqrt{n}(n-1)!}\|f\|_2. \end{align}