How to find the norm of this bounded linear functional? Let $C^\prime[a,b]$ denote the normed space of all continuously differentiable real (or complex) valued functions defined on the closed, bounded interval $[a,b]$ in $\mathbf{R}$ with the norm defined by $$||x||:= \max_{a \leq t \leq b} |x(t)| +  \max_{a \leq t \leq b} |x^\prime(t)| $$ for all $x$ in $C^\prime[a,b]$, and let $f$ be the functional defined by $$f(x):=x^\prime(t_0)$$ for all $x$ in $C^\prime[a,b]$, where $t_0$ is an arbitrary but fixed point in $[a,b]$. Now $f$ is a bounded linear functional. How to compute the norm of $f$, and what is this norm? 
 A: Since for all $x\in C^1[a,b]$ we have $|f(x)|=|x'(t_0)|\leq \|x\|$ it follows that $\|f\|\leq1$.
In order to prove that in fact $\|f\|=1$ we may assume $a\leq0\leq b$ and $t_0=0$. Consider the functions
$$x_n(t):={t\over 1 + n^2 t^2}\ .$$ 
Then 
$$x_n'(t)={1-n^2 t^2\over (1+n^2 t^2)^2}\ ,$$
and it is easy to see that $|x_n'(t)|\leq x_n'(0)=1$ for all $t\in\Bbb R$. Furthermore we can deduce that $|x(t)|$ takes its maximum value ${1\over 2n}$ at $t=\pm{1\over n}$. It follows that for sufficiently large $n$ we have
$$\|x_n\|=1+{1\over 2n}\ ,$$
so that $f(x_n)=x_n'(0)=1$ implies
$$\lim_{n\to\infty}{|f(x_n)|\over\|x_n\|}=1\ ,$$
as claimed.
A: I might be missing something, but try this approach
$$
\left \| f \right \| = \sup_{\left \|x\right\|=1} \left |f(x) \right | = \sup_{\left \|x\right\|=1} \left | x'(t_0) \right | = \sup_{\left \| x \right \|=1}\max \left | x'(t)\right | = \sup_{\left \| x \right \|=1} \left (1-\max\left | x(t)\right |\right) = 1
$$
A: Since $\|x\|=1$ ensures $|x(t_0)|+|x'(t_0)|\leq 1$, then $|x'(t_0)|\leq 1$ so that
$$
\|f\|\leq 1
$$
To show that $\|f\|=1$ it suffices to find a function $x(t)$ s.t. $\|x\|=1$, $x(t_0)=0$ and $|x'(t_0)|=1$, i.e. such that $|f(x)|=1$. Can you define one such $x(t)$ explicitly?
EDIT:
Such a function $x(t)$ can be found by suitably convolving a suitable function: finding one requires a bit of work, but it is trivial that one such exists.
Define
$$
  \xi(t)
~=~
  \textstyle
  {
    \begin{cases}
      -\frac12 & \text{for } ~~ t\in [\,a,\,t_0\,)
    \\
      +\frac12 & \text{for } ~~ t\in [\,t_0,\,b\,]
    \end{cases}
  }
$$
Clearly $\xi\notin\mathcal C'[a,b]$. To make it regular, for $\varepsilon\in(0,1]$ let us define
$$
  x_\varepsilon (t) = \rho_\varepsilon*\xi(t)
$$
where $\rho_\varepsilon$ is a symmetric mollifier required to satisfy some properties we are about to state, and $*$ denotes the convolution. By symmetric we mean that $\rho_\varepsilon(-x)=\rho_\varepsilon(x)$. Now, for any $\varepsilon$ this properties hold:


*

*$x_\varepsilon\in\mathcal C'[a,b]$ (to be more specific, $x_\varepsilon\in\mathcal C^\infty[a,b]$)

*$x_\varepsilon'(t)=\rho_\varepsilon'*\xi(t)$ and $\|x_\varepsilon\|_\infty=|x_\varepsilon'(t_0)|$ (by anti-symmetry of $\rho_\varepsilon'$ and of $\xi$ around $t_0$)

*$\|x_\varepsilon\|_\infty\leq\|\xi\|_\infty=\frac12$;

*$x_\varepsilon(t_0)=0$ (by symmetry of $\rho_\varepsilon$ and anti-symmetry of $\xi$ around $t_0$);


To sum up, you have a smooth function $x_\varepsilon(t)$ with $x_\varepsilon(t_0)=0$ and whose (absolute value of the) derivative has maximum in $t_0$. In particular, by property (2.), $f(x_\varepsilon)=x'_\varepsilon(t_0)=\int_0^\varepsilon \rho_\varepsilon(s)\,{\rm d}s$. Choose $\varepsilon$ so that that quantity equals $1$ and you have $f(x_\varepsilon)=1$ and $\lambda \,\dot=\,\|x_\varepsilon\|=\|x_\varepsilon\|_\infty+\|x_\varepsilon'\|_\infty \geq 1$. (I believe $\|x\|=1$ but I am not sure...)
Shrinking $x_\varepsilon$ vertically by $\lambda$ and horizontally around $t_0$ by $\lambda^{-1}$, i.e. considering
$$
X(t) ~:=~ \lambda x_\varepsilon(t_0+\lambda^{-1}(t-t_0))
$$
gives you $X(t)\in\mathcal C'[a,b]$ with $\|X\|=1$ and $f(X)=1$.
Surely there are simpler ways to find such an $X$...
