Proof of the du Bois-Reymond lemma "by approximation" [closed]

Question

I'm curious about the following argument in Morrey ("Multiple integrals in the calculus of variations", Lemma 2.3.1). Suppose $f\in L^1[0,1]$ and $$\int_0^1 fg\,dx=0$$ for every test function $g\in C^\infty_c[0,1]$. Then, "by approximations," the same is true if $g$ is merely bounded and measurable, in which case we may take $g=\mathrm{sign}(f)f,$ which implies $f=0$ almost everywhere.

How exactly is one supposed to approximate? Certainly not in the $L^\infty$ norm because the closure of $C^\infty_c[0,1]$ in this norm is not $L^\infty$ itself. Furthermore, $\mathrm{sign}(f)f$ need not be in $L^\infty$, so how does this argument work?

Answer 1

One way to run this argument is to note that you can approximate a bounded measurable function by $C^\infty_c$ functions, almost everywhere and boundedly. That is, given a bounded measurable $h$, you can find a uniformly bounded sequence $g_n \in C^\infty_c([0,1])$ with $g_n \to h$ almost everywhere.

There are many ways to see this. For example, look up your favorite proof that $C^\infty_c([0,1])$ is dense in $L^1([0,1])$. So you can find a sequence $g_n \to h$ in $L^1$. Passing to a subsequence, you can get it to converge almost everywhere. Truncating smoothly at a level higher than $\sup |h|$, you can get the $g_n$ to be uniformly bounded.

So then by the dominated convergence theorem, under the given assumptions, you can conclude that $\int f h = 0$ for all bounded measurable $h$. Now take $h = \operatorname{sign} f$, so that $fh=|f|$.

I suspect the reference to $\operatorname{sign}(f) f$ is a typo.

Answer 2

I will prove a slightly more general results.

Theorem. If $f\in L^1_{\rm loc}(\Omega)$, $\Omega\subset\mathbb{R}^n$, and $$ \int_\Omega f\phi=0 \quad \text{for all $\phi\in C_c^\infty(\Omega)$}, $$ then $f=0$ a.e.

Proof. Suppose that $f\neq 0$. We can assume $f$ is positive on a set of positive measure (otherwise we replace $f$ by $−f$). Then there is a compact set $K ⊂ \Omega$, $|K| > 0$ and $ε > 0$ such that $f ≥ ε$ on $K$. Let $G_i$ be a sequence of open sets such that $K ⊂ G_i ⊂⊂ Ω$, $|G_i \setminus K| → 0$ as $i → ∞$. Take $ϕ_i ∈ C^∞_0(G_i)$ with $0 ≤ ϕ_i ≤ 1$, $ϕ_i|_K ≡ 1$. Then $$ 0 = \int_Ω fϕ_i ≥ ε|K| − \int_{G_i\setminus K} |f| → ε|K| , $$ as $i → ∞$, which is a contradiction. The proof is complete.