Counter example to the fundamental lemma of calculus of variations

Question

This is the counter example which my teacher gave $$\int_0^{2\pi}\sin x \sin 2x\:dx=0$$ I tried to disprove him, but I couldn't suceed. Can anyone help me ?

Answer 1

Suppose you give me a continuous function $f$ and ask me if it is identically zero. The fundamental Lemma of calculus of variations tells me that if I take every smooth compactly support function $h$ and show that

$$ \int_D f(x)h(x) \, dx = 0 $$

Then I can conclude $f \equiv 0$. However, what you have done is simply show that for one particular $h$

$$ \int_D f(x)h(x) \, dx = 0 $$

If I take a different $h$, say $h := f$, you will find integral is nonzero. Therefore, $f \not\equiv 0$.

Answer 2

Let us base our argument on the formulation of the lemma in Wikipedia:

If a continuous function $f$ on an open interval $( a , b )$ satisfies the equality $$\tag1\int _{a}^{b}f(x)h(x)\,\mathrm {d} x=0$$ for all compactly supported smooth functions $h$ on $( a , b )$, then $f$ is identically zero.

So in order to present a counter-example (if such existed), we'd have to

• specify an open interval $(a,b)$,
• specify a continuous function $f$ on $(a,b)$,
• show that for all compactly supported smooth functions $h$ on $(a,b)$ equation $(1)$ holds,
• and show that $f$ is not identically zero.

Instead, you merely present a single equation of the form $\int_a^bg(x)\,\mathrm dx=0$. Even when using a crystal ball to interprete that, the closest to a counterexample is that you want $a=0$, $b=2\pi$, $f(x)=\sin x$ (which at least is non-zero), and have verified equation $(1)$ for a single specific case, namely $h(x)=\sin 2x$. However, to qualify as a counterexample this would have to be shown for all allowed choices of $h$.

Abstractly, a counterexmple to a statement of the form $\forall f\colon \Phi(f)$ is a specific $f$ for which $\neg\Phi(f)$ (which thus proves $\exists f\colon \neg \Phi(f)$, the negation of $\forall f\colon \Phi(f)$). Here $\Phi(f)$ is itself of the form $(\forall h\colon \Psi(f,h))\to \Theta(f)$, so $\neg\Phi(f)$ is $(\forall h\colon \Psi(f,h))\land\neg \Theta(f)$; it seems you supposed that $\neg\Phi(f)$ were $(\neg\forall h\colon \Psi(f,h))\to \Theta(f)$ (or equivalently $(\exists h\colon \neg\Psi(f,h))\to \Theta(f)$).