Norm of functional on $C[-1,1]$ defined as $P(f) = f(1)+f(-1)-2f(0)$

Question

I am studying for my final exam and kinda struggling with the following:

Calculate the norm of functional on $C[-1,1]$ defined as $P(f) = f(1)+f(-1)-2f(0)$

This seems like one-liner but I am clueless anyway. I appreciate your time.

Answer 1

Clearly

$$\vert P(f) \vert \le 4 \Vert f \Vert_\infty.$$ So the norm of $P$ is less or equal to $4$. Now $f(x) = -\cos \pi x$ is such that

$$f(0) = -f(1)=-f(-1)=-1$$ and $ \Vert f \Vert_\infty =1$. So $\vert P(f) \vert = 4$ and $\Vert P \Vert = 4$.

Answer 2

Clearly $\|P\| \le 4$. Let $f(x) = 2x^2-1$. Then $$f(1) = f(-1) = 1, \quad f(0) = -1$$ so $P(f) = 4$. On the other hand, we have $$\|f\|_\infty = \sup_{x\in[-1,1]}|f(x)| = \sup_{x\in[0,1]}(2x^2-1) =1$$ so we conclude $\|P\|=4$.