Functional Analysis – Are Smooth Functions with Compact Support Weakly-* Dense in L^??

distribution-theoryfunctional-analysisreference-requesttopological-vector-spaces

My question is this : given $f \in L^\infty(\mathbb{R}^2)$, can we find a sequence $\phi_n$ of smooth, compactly supported functions (test functions) such that for any $g \in L^1(\mathbb{R}^2)$,

$$\int g \phi_n \rightarrow_n \int g f$$

i.e. $\phi_n$ converges weakly to $f$ in the weak * topology of $L^\infty$ ?

I know that a strong convergence is true for $L^p$, $p < \infty$ and wrong for $p=\infty$. However, it seems that if you only ask weak convergence it should be true even in $L^\infty$…

I have not been able to find a reference for this, either in Rudin's Functional Analysis or Brezis's book.

Is this true ? If so, can anyone provide me with a reference ?

Thanks in advance.

Best Answer

It is true. Firstly, it is easy to see that there exists $\{\phi_n\}$, such that for each $n\ge 1$, $\phi_n$ satisfies the following conditions: (i) $\phi_n$ is smooth and compactly supported; (ii) $\|\phi_n\|_\infty\le \|f\|_\infty$; and (iii) $\lim_{n\to\infty} \phi_n=f$ a.e. on $\mathbb{R}^2$. Then the conclusion follows from dominated convergence theorem directly.

Related Question