Extending the Dirac delta to $L^p$

Question

In this question the Dirac delta is extended from $\mathcal C^0([-1,1])$ to $L^\infty([-1,1])$ by the Hahn-Banach theorem.

My question is: why can't it be extended to an arbitrary $L^p([-1,1])$ for $p \geq 1$?

Answer 1

To extend $\delta$ to act on $L^p$ along the idea you mention, you need to consider $C[-1,1]$ as a subspace of $L^p[-1,1]$. In that case $\delta $ is not bounded, so Hahn Banach does not apply.

For instance let $f_n(t)=\max\{0,1-n|t|\}$. Then $\delta(f_n)=1$ for all $n$, while $$\|f_n\|_p=\frac{2^{1/p}}{(p+1)^{1/p}}\,\frac1{n^{1/p}}\to0. $$

Answer 2

$f \to f(0)$ is a continuous linear functional on $C[0,1]$ with the norm from $L^{\infty}$. So Hahn Banach Theorem shows that it extends to a continuous linear functinal on $L^{\infty}$. However, $f \to f(0)$ is not continuous on $C[0,1]$ w.r.t the $L^{p}$ norm. It does not extend to a an element of $L^{p'}=(L^{p})^{*}$.

[Let $f_n(x)=1-nx$ for $0 \leq x \leq \frac 1 n$ and $0$ for $x >\frac 1 n$. Then $f_n$'s are continuous, $f_n \to 0$ in $L^{p}$ but $f_n(0)=1$ for all $n$].