Prove that there is no function $g\in L^1[-1,1]$ such that $f(\phi)=\int_{-1}^1\phi g\,dx$ for all $\phi\in C[-1,1]$

Question

For every $\phi \in C[-1,1]$ define the bounded linear functional $$f(\phi)=\phi(0).$$ Prove that there is no function $g\in L^1[-1,1]$ such that $$ f(\phi) = \int_{-1}^{1} \phi g \, \mathrm{d}x, \qquad \forall \, \phi\in C[-1,1]. $$

I verified that it is of course a bounded function. But to prove the result I used several test functions as $\phi$ but was not successful.

What I was trying to do is that: Since $g \in L^1$, it implies $\int_{-1}^{1} |g| <\infty$. So I thought of getting an expression like $n\leq\int|g|$ by choosing appropriate $\phi_n$ but I couldn't make it.

Answer 1

Let $\phi_n(x)=1-n|x|$ for $|x| \leq \frac 1n$ and $\phi_n(x)=0$ for $|x|>\frac 1 n$. Then we get $1=\phi_n(0)=\int \phi_n(x) g(x)dx$. Use DCT to show that RHS tends to $0$ as $n \to \infty$. This contradiction proves that there is no such $g$.

Note: $\phi_n(x) \to 0$ for every $x \neq 0$ and $0\leq \phi_n(x) \leq 1$ for all $x$.