Let $C[0, 1]$ be the linear space of all continuous functions on the interval $[0, 1]$. Let $δ : C[0, 1] \to \Bbb R$
be the linear functional that evaluates a function at the origin, that is, $δ(f) = f(0)$.
a) Show that if $C([0; 1])$ is equipped with the norm $||f|| = max_{0≤x≤1}|f(x)|$, then $δ$ is
bounded and compute its norm $||δ||$.
b) Show that if $C([0; 1])$ is equipped with the one-norm
$$||f||_1 =\int^1_0 |f(t)| dt $$
then $δ$ is unbounded.
Proof idea (a):
Since $max_{0≤x≤1}|f(x)|=M<\infty$,
$0\leq ||\delta||=sup_{||f||\neq0} \frac{||\delta f||}{||f||}=sup_{||f||\neq0} \frac{max|f(0)|}{max_{0≤x≤1}|f(x)|}=|f(0)|\frac{1}{max_{0≤x≤1}|f(x)|}\leq |f(0)|\frac{2}{M}$ so is bounded.
(b)
$||\delta||=sup \frac{||\delta f||}{||f||}sup \frac{|\int^1_0 |f(0)| dt|}{\int^1_0 |f(t)| dt}=|f(0)|sup\frac{1}{\int^1_0 |f(t)| dt}=…$
Now if I can find a function $f$ such that integrated between 0 and 1 is $F(1)-F(0)=0$, all is well. Or, $f(t)=1/t$ then $F(t) = ln(t)$ and the integral with bounds 0 and 1 is divergent. Am I on the right track?
Thanks and Regards,
Best Answer
You have to come up with continuous functions $f$ , so you cannot use $\frac 1 t$.
If $f_n(x)=1-nx$ for $0 \leq x \leq \frac 1 n$ and $0$ for $\frac 1n \leq x \leq 1$ then $\int_0^{1} |f_n(t)|dt \to 0$ but $\delta (f_n)=1$ for all $n$.