$\int (f-g)\phi^{1/n}=0$ implies $f=g$ ae.

I'm trying to solve a problem that is similar to the fundamental theorem of calculus of variations.

Suppose $\phi$ is a positive, continuous function on $[a,b]$ such that $\phi^{1/n}\to1$ pointwise. Suppose also that $f$ and $g$ are integrable on $[a,b]$ such that for all $n$,

$$\int_{[a,b]}(f-g) \,\phi^{1/n}=0. (*)$$

Show that $f=g$ almost everywhere.

My idea is to show that $\int_{[a,b]}(f-g)^2=0$, from which the result follows. So I have, by Fatou's lemma:

$$\int_{[a,b]}(f-g)^2=\int_{[a,b]}(f-g)^2 \,\lim_{n\to\infty}\phi^{1/n}\leq\liminf_{n\to\infty}\int_{[a,b]}(f-g)^2\phi^{1/n}$$

And then I'm stuck at this point since I'm not sure how to evaluate this liminf. I want to say it is equal to 0, but I can't justify it.

Also, I haven't used above the fact (*). Any hints are appreciated. Thanks.

Solution 1:

The statement is false. Consider $[a;b]=[-1;1]$, $\phi\equiv 1$, $f\equiv 0$ and $g(x)=x$. All the conditions are met, however, $f\neq g$ almost everywhere. Presumably it was meant that the conditions hold for all intervals and not a fixed one.

In fact, for $[a;b]=[-1;1]$, we can pick more generally $\phi \equiv c >0$ and $f(x)=g(-x)$ (and $f$ not even), then we still have a counterexample. And if we wish, we can make the same construction on your favourite compact, nontrivial interval by a suitable change of variables.