Check proof that operator in unbounded please I have to show that $f:\mathcal{C}'[a,b]\rightarrow \mathbb{R}$ with $f(x)=x'(\frac{a+b}{2})$ is unbounded. Here $\mathcal{C}'[a,b]$ (the space of continuously differentiable functions) is to be considered as a subspace of $\mathcal{C}[a,b]$, so with the norm $||x||=\sup_{s}x(s)$.
Now my idea here is that I can take a sequence of functions so that every function has a spike at the centre of the interval and is $0$ everywhere else. The area of this spike should be $1$, and by making the 'base' of the spike smaller as we go along the sequence, this means that it's peak should get higher. However, because the area of the spike remains $1$, if we integrate any function in this sequence, it will have as a maximum value $1$. So we have a sequences $x^{(n)}\in \mathcal{C}[a,b]$ with $||x^{(n)}||=1$, but $f(x)$ get arbitrarily large. Here are the details:
$x'^{(n)}(s)=  
\left\{
     \begin{array}{lr}
       0 & : a\leq s\leq \frac{a+b}{2}-\frac{1}{n}\\
       n^2s-n^2(\frac{a+b}{2}-\frac{1}{n}) & : \frac{a+b}{2}-\frac{1}{n} \leq s \leq \frac{a+b}{2}\\
       -n^2s+n^2(\frac{a+b}{2}+\frac{1}{n}) & : \frac{a+b}{2}\leq s \leq \frac{a+b}{2} + \frac{1}{n}\\
       0 & : \frac{a+b}{2}+\frac{1}{n} \leq s \leq b
     \end{array}
   \right.\\$
These functions are all continuous and $x'^{(n)}(\frac{a+b}{2})=n$
Then the integrals of these functions will be 
$x^{(n)}(s)=  
\left\{
     \begin{array}{lr}
       0 & : a\leq s\leq \frac{a+b}{2}-\frac{1}{n}\\
       \int_{\frac{a+b}{2}-\frac{1}{n}}^{s}n^2s-n^2(\frac{a+b}{2}-\frac{1}{n}) & : \frac{a+b}{2}-\frac{1}{n} \leq s \leq \frac{a+b}{2}\\
       \frac{1}{2}+\int_{\frac{a+b}{2}}^{s}-n^2s+n^2(\frac{a+b}{2}+\frac{1}{n}) & : \frac{a+b}{2}\leq s \leq \frac{a+b}{2} + \frac{1}{n}\\
       1 & : \frac{a+b}{2}+\frac{1}{n} \leq s \leq b
     \end{array}
   \right.\\$
These are also all continuous and $||x^{(n)}||=1$. 
And so finally we can conclude that $f$ cannot be bounded, since it can send an $x\in \mathcal{C}'[a,b]$ with $||x||=1$ to an arbitrarily large $r\in \mathbb{R}$.
I don't really like this proof, because when looking at those piecewise functions here it is far from immediately obvious that this is all correct. This is why I would like to: a) get some verification on the correctness, and b) see an alternative more elegant proof.
Thanks!
 A: Your proof is correct, but as you suspect, is more elaborate than it needs to be. Your idea of taking a sequence of functions is solid though.
Consider the following: without loss of generality, take $a=-1$ and $b=1$. Take
$$
x^{(n)}(t)= \arctan(nt)
$$
A: consider the sequence $x_n(s)=\sin(2\pi(b-a)ns)$, then $x'_n(s)=2\pi(b-a)n\cos(2\pi(b-a)ns)$, here $\|x_n\|=1$, and we see that $$\lim_{n\to\infty}\|f(x_n)\|=\lim_{n\to\infty}2\pi(b-a)n=\infty$$ if you wish to argue with more rigour assume there exists $K\gt 0$ s.t. $\|f(x)\|\le K\|x\|,\,\forall x\in C'(a,b)$, but $$\|f(x_n)\|=2\pi(b-a)n\gt K$$ for sufficiently large $n\in\mathbb N$
A: I agree with our colleague Omnomnomnom; the construction in the question looks fine to me too.  And there's no shame in using the piecewise linear approach; it has many advantages in terms of clarity and ease of use, etc., and these are good things which have been exploited by many mathematicians of great stature; so in using the PL technique, we follow in the footsteps a of a great tradition, also a good thing, methinks.  Of course, specifying all the apparatus needed to desribe a PL function gets a little cumbersome, but the trade-off is that they sure are easy to manipulate once we figure out in just what interval the independent variable lies.  Thinking algorithmically for a moment, we see that we are buying simplicity of evaluation at the cost of searching for the interval in which $t$ lies.  But this is the stuff of computer science, no?
An engaging question is whether there is really any substantive difference in complexity between the PL method and more analytical techniques, based on evaluating more complex functions on fewer intervals.   But that's a tale for telling in some other time, at some other place.
To wrap up my meanderings, here's my candidate for a sequence of functions with the desired properties:  let $m$ (for midpoint) be defined by
$m = \dfrac{a + b}{2}; \tag{1}$
let $\beta_n \in \Bbb R$, $n$ a positve integer, be a positve sequence of reals such that $\beta_n \to \infty$, set
$c_n = (\tanh \beta_n (b - m))^{-1}, \tag{2}$
and finally set
$f_n(t) = c_n \tanh \beta_n (t - m) \tag{3}$
for $t \in [a, b]$.  We have
$f_n'(t) = \dfrac{c_n \beta_n}{\cosh^2 \beta_n (t - m)} > 0; \tag{4}$
thus each $f_n(t)$ is monotonically increasing; furthermore, since
$f_n(a) = c_n \tanh (\dfrac{\beta_n(a - b)}{2}) = -1, \tag{5}$
and
$f_n(b) = c_n \tanh (\dfrac{\beta_n(b - a)}{2}) = 1, \tag{6}$
it is easily seen that $\Vert f_n(t)) \Vert = 1$ for all $n$; the sequence $f_n(t)$ lies entirely in the unit sphere in $\mathcal C[a, b]$.  However, we see from (4) that
$f_n'(m) = \dfrac{c_n \beta_n}{\cosh^2 (0)} = c_n \beta_n \to \infty \tag{7}$
as $n$ increases without bound; this because $c_n \to 1$ from below as $n$ continues to grow, while $\beta_n \to \infty$.  That $c_n \to 1$ monotonically may be seen by inspecting the definition (2) and remembering that $\tanh x \to 1$ monotonically as $t \to \infty$;  that $\beta_n \to \infty$ by our hypothesis.  We thus see that the functional $x(t) \to x'(m)$ maps a sequence in the unit sphere in $\mathcal C[a, b]$ to an unbounded sequence in $\Bbb R$; thus, it cannot be continuous, that is to say, bounded.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
