Show that $(Tf)(x) = \int_0^x f(t)dt$ is bounded

Question

Let $A = C[0,1]$ be the Banach space of all continuous functions on the interval $[0,1]$. Let $T$ be the linear operator from $A$ to $A$ defined by $$(Tf)(x)=\int_0^x f(t)dt.$$

I know that $T$ is linear, and want to show that $T$ is continuous. I must therefore show that $T$ is bounded.

I know that

\begin{align} \Vert Tf \Vert = \max_{ x \in [0,1]} |Tf(x)|. \end{align}

and also

\begin{align} \Vert T\Vert = \sup_{\Vert f \Vert =1} \Vert Tf \Vert = \sup_{f\neq 0} \frac{\Vert Tf \Vert}{\Vert f \Vert}. \end{align}

I think that I need some inequality comparing $\Vert Tf \Vert$ with $\Vert f \Vert$ in order to deduce that $\Vert T \Vert$ is bounded in the above equality. Am I correct, and if so, can someone please point me in the right direction?

Answer 1

For $0 \le x \le 1$ and $f\in C[0,1]$, \begin{align} |Tf(x)| & = \left|\int_0^x f(t)dt\right| \\ & \le \int_0^x|f(t)|dt \\ & \le \int_0^x\|f\|dt \\ & = x\|f\| \le \|f\|. \end{align} Therefore $\|Tf\| =\sup_{x\in[0,1]}|Tf(x)| \le \|f\|$.

Answer 2

You must show that $T$ is bounded, i.e. $$\|Tf\|_{C[0,1]} \le C \|f\|_{C[0,1]}, \quad \forall f \in C[0,1],$$ for some $C > 0$. We have $$|Tf(x)| \le \int_0^x |f(t)|dt \le \|f\|_{C[0,1]}, \quad \forall x \in [0,1],$$ which exactly means $$\|Tf\|_{C[0,1]} \le \|f\|_{C[0,1]}.$$