Computing the norm of operator when space is equipped with sup norm and $L^1$ norm Let $\phi $ be the linear functional $\phi (f)=f(0)-\int^1_{-1}f(t)\:\mathrm{d}t$
a.Compute the norm of $\phi$ as a functional on Banach space $C[-1,1]$ with sup norm.
b.Compute the of $\phi$ as a functional on the normed vector space $C[-1,1]$, equipped with $L^1$-norm
For a. I have done found $\|\phi\|\le3$,is it correct? But i don't which function achieves its maximum
For b. it seems norm is again 3 but not sure,
Please help me..
 A: For (a) we have
$$||\phi||_{op}=\sup_{\{f:||f||_{\infty}=1\}} |\phi(f)|=\sup_{\{f:||f||_{\infty}=1\}}\left|f(0)-\int_{-1}^{1} f(x)dx\right|.$$
The "worst case" is clearly obtained by a function that is $1$ at $x=-1$ and $-1$ for all other $x$.  There is no continuous function with these properties, but one can construct a sequence of continuous functions approaching such a function (non-uniformly of course), whereby which we conclude $||\phi||_{op}=3.$
(b) is handled in the same way, but now consider the worst case with respect to the class of continuous functions with $||f||_{L^{1}([-1,1])}=1$.  As a hint consider a sequence of Dirac like functions $\{f_{n}\}$ concentrated at $x=0$ with total mass $1$ and compute the sequence $|\phi(f_{n})|$.  What can you conclude?
A: b) $\phi $ is not bounded because if we take $f_n : [-1,1]\to \mathbb{R}$ $$f_n (\xi )=\begin{cases} n(1+n\xi ) \mbox{ for } -\frac{1}{n} \leq \xi < 0 \\n(1-n\xi) \mbox{ for } 0\leq \xi \leq\frac{1}{n} \\ 0 \mbox{ for } \frac{1}{n} <|\xi| \leq 1 \end{cases}$$ then $$||f_n ||_{L^1} =1$$ but $$\phi (f_n ) = n -1 .$$
