Let $X=\{f\in C[0,1] :f(0)=0 \}$. Prove that there does not exist $f\in X$ such that $\|f \|_{\infty}\leq 1$ and $|T(f)|= 1$.

Question

Let $X=\{f\in C[0,1] :f(0)=0 \}$. Prove that there does not exist $f\in X$ such that $\|f \|_{\infty}\leq 1$ and $|T(f) |= 1$. Define
\begin{align}T:X&\to \Bbb{R}\\&f\mapsto T(f)=\int^{1}_{0}f(t)dt\end{align}

My trial

Suppose for contradiction that there exists $f\in X$ such that $\|f \|_{\infty}\leq 1$ and $|T(f) |= 1$. Then,

$$1=|T(f) |\leq \sup\limits_{\|f \|\leq 1}|T(f) |=\|T(f) \|.$$

I'm stuck here, I don't see how to get a contradiction. Kindly help.

Answer 1

Let $f\in X$.

You have $f(0) = 0$. By using the continuity of $f$, let $r>0$ be such that for all $x\in [0, r]$, $|f(x)|\le \frac{1}{2}$. Then using the fact that $|f(x)| \le 1$ for all x :

$$ |T(f)| = \left|\int_0^1 f \right| \le \int_0^1 |f(t)|dt = \int_0^r |f(t)|dt + \int_r^1 |f(t)|dt \le \frac{r}{2} + 1-r = 1 - \frac{r}{2}<1$$