Calculating a norm of an operator Let $T \in (C([a, b]))^*$, 
$$ T(u) = \underset{a}{\overset{(a+b)/2}\int} u(x) dx - \underset{(a+b)/2}{\overset{b}\int} u(x) dx. $$
Show that $ || T || = b - a $.
We have that
$$|| T || = \underset{u \in C([a, b]), ||u|| = 1}{sup} | T (u) |$$
I calculated that $$| T(u) | \leq b - a, \forall u \in C([a, b]), ||u|| = 1$$
i.e. $$ ||T|| \leq b - a, $$
but I can't show that  $ b - a \leq ||T|| $. 
$$ | T(u) | = \left| \underset{a}{\overset{(a+b)/2}\int} u(x) dx - \underset{(a+b)/2}{\overset{b}\int} u(x) dx \right| \geq \left| \left| \underset{a}{\overset{(a+b)/2}\int} u(x) dx  \right|- \left| \underset{(a+b)/2}{\overset{b}\int} u(x) dx \right| \right|  $$
and then?
Thank you!
 A: A sketch:
Look at the expression for $T(u)$ and suppose you were allowed to use $u$ defined on $[a,b]$ by $u(x) = 1$ on $[a,(a+b)/2]$ and $u(x) = -1$ on $[(a+b)/2, b]$.  The expression evaluates to $b-a$ here.  Of course, $u$ is not continuous so this does not work for you.
Instead take a sequence of continuous functions that approximate this function - you can even form a particular approximation by a straight line connecting the graph at a bit before the midpoint to a bit after.  You can do some easy estimates to see the integrals converge.  Then you will have that $T(u_n)$ converges to $b-a$, so $\left\| T \right\| \geq b - a$ .
To add a bit of explanation, this is a normal thing to happen when computing operator or functional norms.  There is an "obvious" upper bound that you can infer, so that $\left\| T \right\| \leq M$, but it is not obvious from algebra and calculus that the reverse inequality holds.  
Then the decision tree goes first: is there an element in the space that attains the bound?  A $u$ with $\left\| u\right\| = 1$ so that $|T(u)| = M$?  Or failing that, if you can find an element outside the space where the action of $T$ attains $M$, can you make a sequence in your space that converges to this element?
Here is why a convergent sequence suffices.  You know that $|T(u)| = M$ and that $|T(u_n)| \rightarrow |T(u)| = M$.  So, for any $\epsilon >0$ $|T(u_n)|$ is within $\epsilon$ of $M$ eventually for large enough $n$.  The norm $\|T\|$ is the sup of evaluations at points of unit norm, and therefore for any $\epsilon$, $\| T \| \geq M - \epsilon$ since we have an element that comes that close to $M$.
