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"The fundamental lemma of the calculus of variations states that if the definite integral of the product of a continuous function $f(x)$ and $h(x)$ is zero, for all continuous functions $h(x)$ that vanish at the endpoints of the range of integration and have their first two derivatives continuous, then $f(x)=0$."

Why can't we construct some $h(x)$ that starts at $(a,0)$, has a positive trajectory - imagine, say, an upside down parabola - and then comes back down to end at $(b,0)$, and then some $f(x)$ that is positive for the first half of the interval and negative for the second half (think of some sinusoidal curve)? Then when we integrate, we would basically be adding up a series of positive quantities ($h(x)f(x)$ would be positive), and then a series of negative quantities ($f(x)h(x)$ would be $\text{negative}\times\text{positive} = \text{negative}$). We could then fine-tune to get this to equal zero.

What precisely am I not understanding here?

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What you're missing is that the statement requires that the integral be zero for all functions $h$, not just ones that you construct to match particular trajectories. In particular, an $h$ that's positive on the first half of your interval and negative on your second half will make your integral manifestly positive. – Steven Stadnicki Jul 24 '12 at 21:24

This seems to be a problem with quantifiers.

If I'm understanding you correctly, the fundamental lemma states that:

If $f \in C[a,b]$ is such that $\int_a^b f(x)h(x)\,dx = 0$ for all $h\in C^2[a,b]$ with $h(a) = h(b) = 0$, then $f \equiv 0$.

A counter-example to this claim would mean finding a function $f \in C[a,b]$ with $f \not \equiv 0$ such that $\int_a^b f(x) h(x)\,dx = 0$ for all $h \in C^2[a,b]$ with $h(a) = h(b) = 0$.

In other words: You found a function $f\not \equiv 0$ such that $\int_a^b f(x)h(x)\,dx = 0$ for some function $h \in C^2[a,b]$ with $h(a) = h(b) = 0$. Your function $f$, however, will not satisfy $\int_a^b f(x)h(x)\,dx = 0$ for all $h \in C^2[a,b]$ with $h(a) = h(b) = 0$.

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