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Let $g\in L^{\infty}([0,1])$ (with respesct to Lebesgue measure). Prove that if $$\int_{[0,1]} f(x)g(x) = 0$$ for all continuous functions $f$ on $[0,1]$ then $g(x) =0$ almost everywhere.

I usually do not post without an attempted proof but I am completely lost with this one, a detailed solution should help me but any thing is grealtly appreciated.

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    don't you mean $L^1$ ? – 2017-01-07
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    Can you find a basis in $L^1[0,1]$ which consists of continuous functions only? – 2017-01-07

3 Answers 3

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Clearly, $g\in L^2[0,1]$ as well, and extending $g$ left and right of $[0,1]$, by setting it being equal to zero, we consider $$ g_\delta(x)=\frac{1}{2\delta}\int_{-\delta}^\delta g(x+t)\,dt. $$ Then $g_\delta\in C[0,1]$ and $\|g_\delta-g\|_{L^2}\to 0$, as $\,\delta\to 0$. In particular, $$ 0=\int_0^1 g_\delta(x)\,g(x)\,dx\to \int_0^1 g^2(x)\,dx. $$ Thus $g(x)=0$, a.e.

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    Is this correct? I am not seeing it completely – 2017-01-07
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    I believe the first integral should be on the interval $[x- \delta, x + \delta]$. – 2017-01-07
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    So, is this proof wrong? or are there just minor details that need adjusting? – 2017-01-07
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    Sorry for the misprint. Corrected now. – 2017-01-07
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    We have $\int_0^1 g_{\delta}(x)g(x)dx=0$ because $g_{\delta}$ is continuous and because of the initial hypothesis about $g.$ So $\|g\|^2= $ $\|g\|^2-\int_0^1g_{\delta}(x)g(x)dx= $ $ |\int_0^1 g(x)(g(x)-g_{\delta}(x))dx |$ $\leq \|g\|\cdot \|g-g_\delta\|,$ which $\to 0$ as $\delta \to 0.$ – 2017-01-08
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Every bounded measurable function $b$ on $[0,1]$ is the pointwise a.e. limit of a sequence of continuous functions whose $\sup$-norms are no larger than $\|b\|_\infty.$ So in our problem we can find continuous $f_n(x) \to \text { sgn }g(x)$ a.e., with $\|f_n\|_\infty \le 1.$ The DCT then gives

$$0 = \int_0^1 f_ng \to \int_0^1 (\text {sgn }g)g = \int_0^1 |g|.$$

This implies $g=0$ a.e.

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There is a sequence $f_n\in C([0,1])$ such that $f_n\to f$ in the $L^1$ norm. Then, applying the DCT, we have $0=\lim \int f_n\cdot gdx\to \int f\cdot gdx$.

But now, the result follows because the map $g\mapsto \phi_g$ with $\phi_g (f)=\int f\cdot gdx$ is an isometric isomorphism of $L^{\infty}$ onto $(L^1)^*.$