Finding the norm of linear functional f(x)

Question

I've got this linear functional $f(x)=x(0)$ where $x=x(t)\in C[-1,1]$.

Should show that it is bounded and calculate its norm.

I tried to do like this $|x(0)|\leq \|x\|_{C[-1,1]} $

$|f|=\sup_{\|x\|_{C[-1,1]}=1} |f(x)| \leq \sup_{\|x\|_{C[-1,1]}=1} |x(0)|$

So norm of $f$ will be less or equal than this $$\sup_{\|x\|_{C[-1,1]}=1} |x(0)|\leq \sup_{\|x\|_{C[-1,1]}=1} \|x\|_{C[-1,1]}=1.$$

And Finally can I take $x(t)=1$ and prove that $|f|=1$?

Thanks in Advance.

Answer 1

You need to prove two things:

1. $\lVert f \Vert \leq 1$. This follows from observing that $|f(x)| = | x(0) | \leq \lVert x \rVert$ and the infimum definition of norm.

2. For some $x \in C[-1,1]$, $| f(x) | = \lVert x \rVert$. This part can be proved by an example of $x = 1$.

Answer 2

Your proof is correct but can be expressed a little more succinctly: $|f(x)| = |x(0)| \le \|x\|$. It follows from this that $\|f\| \le 1$. Taking $x=1$ shows that the bound is attained.