If $T:C[0,1] \rightarrow \mathbb{R}$ is defined by $T_{x_0}(f)=f(x_0)$ then $||T||=1$.

Question

Let $x_0 \in [0,1]$. Define $T_{x_0}:C[0,1] \rightarrow \mathbb{R}$ by $T_{x_0}(f)=f(x_0)$.

$||T||:= \sup\{|T_{x_0}(f)|: ||f||_{\infty} \leq 1\}$ where $||f||_{\infty}:=\max\{|f(x)|:x \in [0,1]\}$.

Prove that $||T||=1$.

For every $f \in C[0,1]$, if $||f||_\infty \leq 1$ then $|f(x)| \leq 1$ for all $x \in [0,1]$. In particular, $|f(x_0)| \leq 1$. So the set $\{|T_{x_0}(f)|: ||f||_{\infty} \leq 1\}$ is bounded above by $1$.

Assume some $a$ such that $0 \leq a <1$ is the least upper bound. How can I derive a contradiction?

Answer 1

Suppose that $f$ is the constant function $1$. Then $\lVert f\rVert=1$ and $\bigl\lvert f(x_0)\bigr\rvert=1$. So, $\lVert T_{x_0}\rVert=1$.