Define the linear functional:
$$f(x)=\int_{-1}^{0}x(t)dt-\int_0^1x(t)dt$$
On the normed space $C[-1,1]$ which consists of all contiuous functions on the interval. The norm is defined as: $\|x\|= \max_{t\in[-1,1]}|x(t)|$
Now, let $x(t)\in C[-1,1]$ be arbitrary
$$\left| f(x)\right|=\left| \int_{-1}^{0}x(t)dt-\int_0^1x(t)dt\right|$$
$$|f(x)|\leq \left|\int_{-1}^{0}x(t)dt\right|+\left|\int_0^1x(t)dt\right|$$
Also,
$$\left|\int_{-1}^{0}x(t)dt\right|
\leq (0-(-1))\cdot \max_{\ t\in[-1,1]}|x(t)|=\|x\|$$
In a similar way for the other integral, one can conclude:
$$|f(x)|\leq 2\|x\|\implies \frac{|f(x)|}{\|x\|}\leq 2$$
Excluding the case where $x(t)$ is the zero function.
Define $\|f\|$:
$$\|f\|=\sup_{x\in C[-1,1]\setminus\{0\}} \left( \frac{|f(x)|}{\|x\|}\right) $$
So $\|f\|\leq 2 $ since $x(t)$ was arbitrarily chosen and we've established an upper bound. But we can't conclude $\|f\| = 2 $ from this since it need not be the least upper bound.
It would be sufficient at this point to show $\frac{|f(x)|}{\|x\|}\geq 2$ for some specific $x(t)\in C[-1,1]$.
I'm having no luck finding this $x(t)$. What's clear is that it needs to be some anti-symmetric function on the interval that attains a maximum value of $1$.
Any suggestions would be appreciated.
Best Answer
I don't think you can find $x$. the idea is that the perfect $x$ would be the function thqt has value -1 on $[-1,0[$ and 1 on $[0,1]$. The problem is that this function is not continuous...
Never mind! try to approximate it with well chosen partially linear functions. You don't need to find an $x\in C[-1,1]$ such that $\frac {\vert f(x)\vert}{\Vert x \Vert}\geqslant 2$.
It suffices to prove that, for any $\epsilon>0$, you can design a continuous function $x$ such that $\frac {\vert f(x)\vert}{\Vert x \Vert}\geqslant 2-\epsilon$.