Changing variable in a supremum integral inequality

Question

So I'm trying to prove that this operator $$ T_n(f)=\int_0^t \cfrac{(t-x)^n}{(n-1)!}f(x)dx $$ is continuous and then calculate its operator norm, provided that

• the operator is a map $E\to E$ with $E=C([0,1],\Bbb R)$, and
• the norm considered on $E$ is the infinite norm.

To prove the continuity I tried to find a $M>0$ such that $\|T_n(f)\|_\infty < M\|f\|_\infty $.

After developing I got to this inequality : $$ ||T_n(f)||_\infty < \int_0^t\sup_{t\in[0,1]} \cfrac{(t-x)^n}{(n-1)!}f(x)dx $$

What I'm trying to do is change the $t$ of the supremum to $x$ so that I could sort the $f(x)$ out and integrate the quantity $\cfrac{(t-x)^n}{(n-1)!}$ to get the required $M$ for the continuity.

Any tips on how to proceed?

Answer 1

Observe that $$ \int_0^t \cfrac{(t-x)^n}{(n-1)!} dx= \cfrac{t^n}{n!} \,. $$ Therefore

$$\|T_n(f)\|_\infty \le \cfrac{1}{n!} \|f\|_\infty \,,\quad (*)$$ with equality for $f \equiv 1$, so the operator norm of $T_n$ is $\cfrac{1}{n!}$.