Let $\delta$ be a linear functional equipped with the sup-norm. Show that $\delta$ is bounded and compute its norm.

Question

Let $\delta:C([0,1])\rightarrow\mathbb{R}$ be the linear functional at the origin: $\delta(f) = f(0)$. If $C([0,1])$ is equipped with the sup-norm $$\|f\|_{\infty} = \sup_{0\leq x\leq 1}|f(x)|.$$ Show that $\delta$ is bounded and compute its norm.

I have this solution:

$|\delta(f)| = |f(0)|\leq \sup_{0\leq x\leq 1}|f(x)| = \|f\|_{\infty}$.

So, $\dfrac{|\delta(f)|} {\|f\|_{\rlap\infty}}\leq 1$

Let $f = 1$ then $|\delta(f)| = \|f\|_{\infty} = 1$. Also $\|\delta\| \geq \dfrac{|\delta(f)|}{\|f\|_{\rlap\infty}} = 1$.

My questions are why set $f = 1$?

Where does the inequality $\|\delta\| \geq \dfrac{|\delta(f)|}{\|f\|_{\rlap\infty}}$ come from?

Thank you for any help and comments in advance.

Answer 1

One of equivalent definitions of the norm of a functional $\delta$ is $$\sup_{f\ne 0} \frac{|\delta(f)|}{\|f\|}$$ (Another common version is to restrict the supremum to the set with $\|f\|=1$, in which case the denominator is not needed).

To determine the value of supremum, we do two things:

(a) establish an upper bound;

(b) show than nothing smaller than it is an upper bound.

In this example, (a) consists of showing $|\delta(f)|/ \|f\| \le 1$ for all $f$; part (b) is done by exhibiting $f$ such that $|\delta(f)|/\|f\|=1$.