Let $\|x\|= max|x(t)|+max|x'(t)|$ be a norm on $C'[a,b]$. Let $f(x)=x'(a/2+b/2)$ be a bounded linear functional on $C'[a,b]$. Find $||f||$. Let $\|x\|= max|x(t)|+max|x'(t)|$ be a norm on $C'[a,b]$. Let $f(x)=x'(a/2+b/2)$ be a bounded linear functional on $C'[a,b]$. Find $\|f\|$.
I got $|f(x)| \leq \|x\|$ for all $x$, and I tried to find an $x(t) \in C'[a,b]$ with $\|x\|=1$ and $|f(x)|=1$ so that we can say $\|f\|=1$ but I got that there is no any such an $x.$ So now I am trying to prove that for each $\epsilon>0$ there exists an $x \in C'[a,b]$ with $\|x\|=1$ such that $|f(x)|>1-\epsilon.$ But could not prove. Please give some hint/way.
 A: Clearly, $\|f\|\leq 1$, as you have already pointed out. For $n\in \mathbb{N}$, consider $$x_n(t) = \frac{\sin(\alpha n t)}{1+\alpha n}$$ where $\alpha > 0$ is chosen so that $\frac{(a+b)\alpha}{4\pi}$ is irrational. For $n$ large enough, $$\max\lvert x_n(t)\rvert = \frac{1}{1+\alpha n}$$ and $$\max\lvert x_n'(t)\rvert = \frac{\alpha n}{1+\alpha n}$$ so $\|x_n\| = 1$. Then, $$x_n'\left(\frac{a+b}{2}\right) = \frac{\alpha n}{1+\alpha n}\cos\left(\frac{(a+b)\alpha n}{2}\right)$$ As $\frac{(a+b)\alpha}{4\pi}$ is irrational, we have by Weyl's equidistribution theorem that $\frac{(a+b)\alpha n}{2}\bmod 2\pi$ is uniformly distributed in $[0, 2\pi)$. Therefore, we choose some increasing $\{n_k\}_{k=1}^{\infty}$ such that $$\left(\frac{(a+b)\alpha n_k}{2}\bmod 2\pi\right)\to 0$$ as $k\to \infty$. Then, $\frac{\alpha n_k}{1+\alpha n_k}\to 1$ and $\cos\left(\frac{(a+b)\alpha n_k}{2}\right)\to 1$, so $$x_{n_k}'\left(\frac{a+b}{2}\right) = \frac{\alpha n_k}{1+\alpha n_k}\cos\left(\frac{(a+b)\alpha n_k}{2}\right)\to 1$$ as $k\to \infty$. Therefore, $\|f\| = 1$.
A: Following the idea from @Jouy Zou's comment, let's try to find a function $g_\varepsilon \in C^1[a,b]$ such that $\lVert g_\varepsilon\rVert_\infty \leq \varepsilon$ and $|f(g_\varepsilon)|\geq 1-\varepsilon$, and such that $\lVert g_\varepsilon\rVert = 1$.
Consider the general Gaussian $\phi(x) = A\,\mathrm{e}^{-\frac{\left(x-\mu\right)^2}{2\sigma^2}}$ centered at $\mu$. Its maximum value is $A$ and the maximum is attained at $\mu$.
The derivative of this Gaussian is 
$$\phi'(x)=A\,\mathrm{e}^{-\frac{\left(x-\mu\right)^2}{2\sigma^2}}\frac{1}{\sigma^2}\left(x-\mu\right)$$
After calculating $\phi''(x)$ and equating with zero we get that $\phi'$ attains its extremal values at $\mu\pm\sigma$, which are $\pm \frac{A}{\sigma\sqrt{\mathrm{e}}}$.
Now we see that if we wish $\lVert \phi\rVert_\infty = \varepsilon$, $\lVert \phi'\rVert_\infty = |f(\phi)|= 1-\varepsilon$, we should center the Gaussian so that the maximum of $|\phi'|$ is $1-\varepsilon$, it is attained exactly at $\frac{a+b}{2}$, and the maximum of $\phi$ is $\varepsilon$.
That's how we come up with:
$$g_\varepsilon(x)=\varepsilon\,\mathrm{e}^{-\frac{\left(x-\frac{a+b}{2}+\sigma\right)^2}{2\sigma^2}}, \forall x \in [a, b]$$
where $\sigma = \frac{\varepsilon}{(1-\varepsilon)\sqrt{\mathrm{e}}}$.
You can verify that $|f(g_\varepsilon)| = 1-\varepsilon$ and $\lVert g_\varepsilon \rVert = 1$.
Since we can do this for every $\varepsilon > 0$, we have $\lVert f \rVert = 1$.
