Find the norm of S

Question

Given $C[0,2]$ with the the max-norm. Let $X= \{x\in C ([0,2]): x (1) =0\}$. We define $S:X\to\mathbb{R}$ as $S(x)=\int_{0}^{2}x(t)dt$ . Compute $||S||$. I have already found that $||S||\leq2$, but I haven't come up with the way to prove $||S||\geq 2$. Are there any hints?. Thanks in advance.

Answer 1

I thought I would spell out in some more detail the point discussed in the comments by me and other contributors. We all more or less suggest the same thing.

The point is that you can't just choose $x(t) = 1$ for every $t$ since you have the additional constraint that $x(1)=0$. But you can see by sketching graphs that you could almost have a function which is $1$ everywhere, except for some small interval around $t=1$.

Remember that $$\|S\|_* = \sup_{\max x(t) =1}|Sx|.$$ You know that $$\|S\|_*\leq 2,$$ and it seems plausible that we actually have equality. Suppose for a contradiction that we did not have equality. Then there exists $\alpha<2$ such that $\alpha$ is an upper bound for the set $\{|Sx| : \max x(t) =1\}.$ If we call $\alpha = 2-\epsilon$ for some $\epsilon>0$, it suffices to show that there exists $\tilde x\in X$ such that $$|S\tilde x|\geq 2-\epsilon.$$

For this, I suggested choosing the function $$\tilde x(t) = \begin{cases}-\frac{2}{\epsilon} t+\frac{2}{\epsilon} & \text{ if } 1-\frac\epsilon 2<t\leq 1\\ \frac 2\epsilon t-\frac 2\epsilon &\text{ if }1<t\leq 1+\frac\epsilon 2\\ 1&\text{ otherwise.}\end{cases}$$

See pictured with $\epsilon = 0.3$. Then the integral is $\int_0^2 \tilde x\, dt = 2- \frac 12 \epsilon >2-\epsilon.$ This leads us to a contradiction. Alternatively you could have chosen $\tilde x$ as Ryszard Szwarc suggested, as some function $\tilde x(t) = |t-1|^{1/n}$ where you would need to find some $n$ sufficiently large so that a similar estimate for the integral holds.