# Show that functional $f(x)=x'((a+b)/2)$ is bounded in $C^1([a,b])$ but unbounded in $C([a,b])$

Let $C^1([a,b])$ be normed vector space of all continuously differentiable functions defined on $[a,b]$ with a norm $$||x||=\sup_{t \in [a,b]} |x(t)|+\sup_{t \in [a,b]} |x'(t)|$$ We have to show that the functional $f(x)=x'((a+b)/2)$ is bounded on $C^1([a,b])$ and not bounded on $C([a,b])$ (space of continuously differentiable functions with a norm $||x||=\sup_{t \in [a,b]} |x(t)|$.

Showing that $f(x)=x'((a+b)/2)$ is bounded on $C^1([a,b])$ is easy here is the proof \begin{align} |f(x)|=|x'((a+b)/2)| \le sup_{t \in [a,b]} |x'(t)| \le sup_{t \in [a,b]} |x'(t)|+\sup_{t \in [a,b]} |x(t)|=||x|| \end{align}

But how to show that it's unbounded on $C([a,b])$? Also, is this particular functional important in some way?

• What is $x'$ in $\mathcal C^0([a,b])$ ? Like is written, $f$ is not defined on $\mathcal C^0([a,b])$.
– idm
Commented Sep 25, 2015 at 13:28
• I said let $C([a,b])$ be a space of continuously differentiable function so it's all most the same as $C^1([a,b])$ but the norms are defined differently.
– Boby
Commented Sep 25, 2015 at 13:31

To show this, find a sequence of continuously differentiable functions with norm $\leq 1$, i.e $x_n\in C^1([a,b]):\|x_n\|_\infty:=\sup\limits_{t\in[a,b]}{|x_n(t)|}\leq 1$, such that $|x_n'(\frac{a+b}{2})|\to\infty$. For example something like the $\arctan$-function, but getting steeper at $t=\frac{a+b}{2}$:
$x_n(t)=\frac{2}{\pi}\arctan (n t)$ in the interval $[a,b]=[-1,1]$. You have $|x_n'(0)|=\frac{2}{\pi}n\to\infty$.
It suffices to show that there are functions $x$ with norm $||x||\leq1$ but with $|f(x)|$ arbitrarly large.
For instance, you can try with $x(t) = \sin(\omega t + \phi)$, for a good choice of $\omega$ and $\phi$. Since $x(t)=\sin(\dots)$, $||x|| \leq 1$. Using a big enough $\omega$ will cause the function to greatly oscillate, and adjusting $\phi$ can make the oscillation occur at $t = \frac{a+b}{2}$.
More explicitely, $\omega \geq M$ and $\phi = - \omega\frac{a+b}{2}$ yields $|f(x)|\geq M||x||$