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

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!!
 A: 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 $T^n$.

[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.

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? :-)
A: 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
$$\|K\|_{L^1(0,1)}=\frac{\int_0^1r^{n-1}dr}{(n-1)!}=\frac1{n!}.$$
A: 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 $$\|T^{n}\|\leq \dfrac{1}{\sqrt{n}(n-1)!}.$$
A: 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}
