I am given a functional $F:C([0,1],\|x\|_\infty) \to \mathbb C $ by formula

$$F(f)=2\int_0^{1/2}f(t)dt \text{ - } \int_{1/2}^1f(t)dt$$

I should find its operator norm. I've found that $\|F\| \le 3/2$ using some basic inequalities, but am unable to prove the equality here, not to mention that I could be wrong and $3/2$ is not the optimal bound.

Any help?


Approximate the function $$ f(x)=\begin{cases}1,&x\leq 1/2\\ -1,&x>1/2 \end{cases} $$ with continuous functions.

For this, we take the sequence of norm $1$ continuous functions: $$ f_n(x)=\begin{cases} 1,&x\leq 1/2-1/n\\ -n(x-1/2),&1/2-1/n<x\leq 1/2+1/n\\ -1,&1/2+1/n<x\leq 1 \end{cases} $$ I leave to you to verify $F(f_n)\uparrow 3/2$

  • 2
    $\begingroup$ Brilliant, the idea is pretty clear and applicable in other examples. Thanks. $\endgroup$ – windircurse Nov 13 '18 at 18:32

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