# 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..

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 $$\phi (f_n ) = n -1 .$$

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?

• I am sorry,are you sure $f=-1$ gives $3$? my computation gives 1 only – Toeplitz Jun 22 '14 at 21:04
• i did that for b) on similar fashion but i ended up with $\int^1_{-1}f(0)+||f||_1$ after substituting $||f||_1=1$ i get only $2f(0)+2$ how it would be $1$, please do full solution for second one – Toeplitz Jun 22 '14 at 21:09
• See my edit for (a). I made a slight oversight and you are correct about your comment. – Sargera Jun 22 '14 at 21:10
• I'm not sure why you're being asked such a question if you don't follow (b). Dirac functions to which I am referring are nonnegative bump functions that spike near the origin and approach 0 elsewhere (rapidly), but such that their total integral is always $1$. In particular, they belong to the class of functions you must consider in evaluating the $\sup$ appearing in the definiton of $||\phi||$. – Sargera Jun 22 '14 at 21:22
• For the example sequence of continuous functions of $L1$ norm $1$ we have $|\phi(f_{n})|\to\infty$ as $n\to\infty$ by construction of the Dirac functions outlined above. Do you see this at least? Therefore, since $||\phi||\geq|\phi(f_{n})|$ for all $n$, the claim follows. – Sargera Jun 22 '14 at 21:28