Show that $\|T^n\|=\frac{1}{n!}$ for $(Tf)(x)=\int_0^xf(t)dt, f\in C[0,1]$

Question

Let $C[0,1]$ and define, $$ (T f )(x) =\int_0^x f (t) dt, \ x ∈ [0, 1]$$ for $f ∈ C[0, 1]$. Show that $\|T^n\|=\frac{1}{n!}$

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By definition, I know that $\|T\|=\sup_{\|f\|=1}\|Tf(x)\|$, using the norm of $C[0,1]$, we have that $$\begin{aligned} \|T\| &= \left\Vert\int_0^x f(t)dt\right\Vert \\ &\leq \int_0^x \|f(t)\|dt =t \\ &\leq 1 \end{aligned}$$

And since I can find a function whose norm is 1 (any function such that $f(x)=1$, for some $x\in[0,1]$). I know that $$\sup_{\|f\|=1}\|(Tf)(x)\|=1$$ (Couldn't find a more elegant way to conclude that).

I suppose then I'll have to use an induction argument, however, I'm struggling finding a way to prove it for another $n$. I'll have that, $$\|T^2f\|=\left\Vert\int_0^t\int_0^tf(t)dt\right\Vert$$ But I'm concluding this is one again by the same arguments I had for $T$. What am I not seeing?

Answer 1

Hints:

• Show by induction that for all $n\geq 0$ and all $x\in[0,1]$, $|(T^nf)(x)|\leq\frac{x^n}{n!}\|f\|$. What does this tell you about $\|T^n\|$?
• Let $f_0=1$ be the constant function with value $1$, and use induction to compute $T^n(f_0)$ for each for each $n\geq 0$. Therefore…