Compute $\|T\|$ when $D (T)$ is endowed with $\|\cdot \|_{\infty}$ norm

Question

Let $X=\{f\in C[0,1]:f (0)=0\}$. Define $$T:X \to \mathbb R$$ $$T_f=\int_0^1 f (t)dt$$ Compute $\|T\|$ when $X$ is endowed with $\|\cdot\|_{\infty}$.

By approximating, I have $$|T_f| \leq\int_0^1 |f (t)|dt \leq \sup_{t\in [0,1]} |f (t)| \int_0^1 dt = ||f||_{\infty}$$ Thus, $\|T\| \leq 1$.

My challenge is to show $\|T\| \geq 1$. To do this I have to find an $f \in X$ such that $\|f\| \leq 1$ and $|T_f| = 1$. Can someone help. Thanks

Answer 1

Such an $f$ won't exist. You would like to take $f(x) = 1$ for all $x \in [0, 1]$, but we require $f(0) = 0$. Hence we approximate $f$ by a sequence of $f_n \in X$, in particular something like

$$ f_n(x) = \begin{cases} nx & 0 \leq x \leq \frac{1}{n} \\ 1 & x \geq \frac{1}{n} \end{cases}$$

You can show $\lvert T(f_n) \rvert \to 1$ which will show that $\lVert T \rVert \geq 1$.