Induction relation of a linear operator

Question

Let $X=C[0,1]$ with the usual norm. Let \begin{equation*} \begin{split} K: X & \to X \\ x(t) &\mapsto K_x(t):=\int_{0}^t x(\tau)d\tau \end{split} \end{equation*} for $x\in X$ and $t\in[0,1]$. The norm we wil use is $\|x\|=\max\{|x(t)|:t\in[0,1]\}$

$(1)$ Prove that $$|K^n_x(\tau)|\leq \frac{\tau^n}{n!}\|x\|$$ for each $\tau\in[0,1]$ and for any $x\in X$.

$(2$) Show that the linear operator $I-K$ is injective.

For the first part, I proceed by induction.

If $n=1$, then $$|K_x(\tau)|=\left|\int_{0}^\tau x(t)dt\right|\leq (\tau -0) \max\{|x(t)|:t\in[0,1]\}= \frac{\tau}{1!}\|x\|$$ However, I don't know how to approach the general case.

For the second part, I know that if $V$ is a Banach space and $A$ is a bounded linear operator with $\|A\|<1$, then $I-A$ is invertible (in particular, injective). However, if the equation in $(1)$ is true, then $$||K^n||\leq \frac{1}{n!}, \quad n=1,2,\ldots$$ and in the case $n=1$ we get that $||K||=1$.

Any help would be appreciated.

Answer 1

(1) $|K^{n+1}_x(t)| \le \int_0^t |K^n_x(\tau)| d \tau \le \int_0^t \frac{\tau^n}{n!}\|x\| d \tau.$ Can you proceed ?

(2) Let $x \in X$ and $Kx=x$. This means $\int_{0}^t x(\tau)d\tau = x(t)$ for all $t \in [0,1].$ This shows that $x$ is differentiable and differentiation gives: $x(t)=x'(t)$

Hence $x(t)=ce^t$ for some constant $c$. Now use again the equation $Kx=x$ to see that $c=0.$

Answer 2

If $\bigl\lvert K_x^n(\tau)\bigr\rvert\leqslant\dfrac{\tau^n}{n!}\lVert x\rVert,$ then\begin{align}\bigl\lvert K_x^{n+1}(\tau)\bigr\rvert&=\left\lvert K\bigl(K_x^n\bigr)(\tau)\right\rvert\\&=\left\lvert\int_0^\tau K_x^n(t)\,\mathrm dt\right\rvert\\&\leqslant\int_0^\tau\bigl\lvert K_x^n(t)\bigr\rvert\,\mathrm dt\\&\leqslant\int_0^\tau\frac{t^n}{n!}\lVert x\rVert\,\mathrm dt\\&=\frac{\tau^{n+1}}{(n+1)!}.\end{align}

In order to prove that $\operatorname{Id}-K$ is injective, just prove that $\ker(\operatorname{Id}-K)=\{0\}$.