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Let $C[-1,1]$ denote the normed space of all (real or complex-valued) functions defined and continuous on the closed interval $[-1,1]$ on the real line, with the norm given by $$\Vert x \Vert_{C[-1,1]} \colon= \max_{0 \leq t \leq 1} \vert x(t) \vert \ \ \ \mbox{ for all } \ \ x \in C[-1,1].$$

Let the functional $f$ on $C[-1,1]$ be defined by $$f(x) \colon= \int_{-1}^0 x(t) \ \mathrm{d} t \ - \ \int_0^1 x(t) \ \mathrm{d} t \ \ \ \mbox{ for all } \ \ x \in C[-1,1].$$

Then $f$ is linear and bounded with norm $\Vert f \Vert \leq 2$. How to determine the exact value of $\Vert f \Vert$?

My work:

For any $x \in C[-1, 1]$, we have \begin{eqnarray*} \vert f(x) \vert &=& \left\vert \int_{-1}^0 x(t) \ \mathrm{d} t \ - \ \int_0^1 x(t) \ \mathrm{d} t \ \right\vert \\ &\leq& \left\vert \int_{-1}^0 x(t) \ \mathrm{d} t \ \right\vert \ + \ \left\vert \int_0^1 x(t) \ \mathrm{d} t \ \right\vert \\ &\leq& \int_{-1}^0 \vert x(t) \vert \ \mathrm{d} t + \int_0^1 \vert x(t) \vert \ \mathrm{d} t \\ &=& \int_{-1}^1 \vert x(t) \vert \ \mathrm{d} t \\ &\leq& \int_{-1}^1 \max_{0\leq s \leq 1} \vert x(s) \vert \ \mathrm{d} t \\ &=& \int_{-1}^1 \Vert x \Vert_{C[-1, 1]} \ \mathrm{d} t \\ &=& \Vert x \Vert_{C[-1, 1]} \ \int_{-1}^1 \mathrm{d} t \\ &=& 2 \Vert x \Vert_{C[-1,1]}. \end{eqnarray*} Thus, $f$ is bounded and taking the supremum over all $x \in C[-1, 1]$ such that $\Vert x \Vert_{C[-1, 1]} = 1$, we get $$\Vert f \Vert \leq 2.$$

Now we require an element $x_0 \in C[-1, 1]$ such that $\Vert x_0 \Vert_{C[-1, 1]} = 1$ and $\vert f(x_0) \vert = 2$.

However, I haven't been able to locate such an $x_0$. Can anybody please be of any help?

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  • $\begingroup$ As Batman's answer shows, you don't "require an element $x_0\in C[-1,1]$" that attains the supremum in the definition of norms of functionals; you only need to get arbitrarily close to the supremum. That's why it's a sup, not a max. $\endgroup$ – Andreas Blass Apr 4 '15 at 14:10
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Look at functions which are $-1$ on $x<- \epsilon$, $1$ on $x > \epsilon$ and go linearly from $-1$ to $1$ on the interval $[-\epsilon,\epsilon]$.

Making $\epsilon$ arbitrarily small gives a function which is still in $C[-1,1]$ with norm $1$ and the value of $f$ for these is $2-\epsilon^2$.

So, $||f|| \geq 2 - \epsilon^2$ for any $\epsilon >0$. Combine this with your upper bound and you're done.

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  • $\begingroup$ let's take $0<\epsilon<1$, and let $x(t)=-1$ if $-1\leq t\leq -\epsilon$; $x(t) = t/\epsilon$ if $-\epsilon\leq t\let\epsilon$; $x(t)=1$ if $\epsilon\leq t\leq 1$. Then $\Vert x \Vert =1$. And, $$\vert f(x) \vert = \left\vert \int_{-1}^{-\epsilon} \ (-1) \ \mathrm{d} t + \int_{-\epsilon}^0 \ \frac{t}{\epsilon} \ \mathrm{d} t - \int_0^{\epsilon} \ \frac{t}{\epsilon} \ \mathrm{d} t - \int_{\epsilon}^1 \ 1 \ \mathrm{d} t \right\vert = \left\vert \epsilon-1 - \frac{\epsilon}{2} - \frac{\epsilon}{2} - 1 + \epsilon \right\vert = \left\vert \epsilon - 2 \right\vert = 2-\epsilon.$$ Right? $\endgroup$ – Saaqib Mahmood Jun 23 '16 at 15:42

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