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Lemma 2 of the Calculus of Variations

I''m trying to understand the lemma but I don't understand why is that $h(x)$ is defined as $$\int_a^x [\alpha(\xi)-c]d\xi$$

In other words, I don't get the point why is that $h(x)$ is defined based on the function $\alpha$ if $h(x)$ can be any function.

How can I know that with that function, $h(x)$ can be expressed any function derivable in $[a,b]$?

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    $\begingroup$ You just pick convenient $h$ to get the result: $$\int_a^b (\alpha(x)-c)^2\, dx = \int_a^b (\alpha(x)-c)h'(x)\, dx = 0$$ implies that $\alpha(x)-c = 0$. $\endgroup$ – Ennar Aug 21 '17 at 4:37
  • $\begingroup$ But the lemma is based in every $h(x)$ with that conditions, I don't get the point how defining conveniently in that form you can express every function $h(x)$ $\endgroup$ – Joan J. Cáceres Aug 21 '17 at 4:56
  • $\begingroup$ I don't know what is troubling you, we don't want to prove anything about $h$'s, we want to prove something about $\alpha$. If all plumbers can fix water pipes, I don't have to call every plumber in the world when my sink is not working, I can just call my neighbor Luigi the Plumber to fix it. If the analogy is not clear enough, "$\alpha$ is constant" is my broken sink that needs fixing, $h$'s are plumbers who can fix it, $h(x) = \int_a^x(\alpha(\xi)-c)\,d\xi$ is my neighbor Luigi who I call to do the job. $\endgroup$ – Ennar Aug 21 '17 at 9:17
  • $\begingroup$ My trouble is that it seems that defining in that way $h(x)$ and following the deduction, the proof of $\alpha$ is just based in a specific type of $h(x)$. For me, by now, is like trying to demonstrate the Lemma, for example, just when $h(x)$ is a constant o a linear function... $\endgroup$ – Joan J. Cáceres Aug 21 '17 at 11:33
  • $\begingroup$ What about my analogy doesn't work for you? It is not specific type of $h$, it is one particular $h$ out of infinitely many. The assumptions are "for all $h$ such and such, this formula is true", the proof is "ok, if it works for all $h$, then it works for this $h$", and finally the conclusion is "look, this one of the many $h$'s I could have chosen helps me prove the lemma". I think you are confusing what are assumptions and what is to be proven. $\endgroup$ – Ennar Aug 21 '17 at 12:11
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If $\alpha(x)$ is not $c$ all the time, then $\alpha(x) - c$ is not zero all the time. An easy way to detect this is to square it and integrate. Conveniently, by the fundamental theorem of calculus, the given $h(x)$ satisfies $h'(x) = \alpha(x)-c$. So $\int_a^b (\alpha(x) -c)h'(x) \,\mathrm{d}x$ is squaring it and integrating (which is what the last display in your imgur link says).

The integral is also like a dot product. One way to get a large dot product from a nonzero vector is to dot the vector with itself. This choice of $h$ produces an integral that is the dot product of $\alpha(x) -c$ with itself. So if $\alpha(x) -c$ is not "the zero vector", we get a "large" (positive) dot product. Since we can distribute $h'(x)$ and actually do the integrals ("on the one hand") getting zero, we end up showing that the dot product is zero, so the vector is zero, so $\alpha(x) = c$.

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  • $\begingroup$ But the lemma is based in every $h(x)$ with that conditions, I don't get the point how defining conveniently in that form you can express every function $h(x)$ $\endgroup$ – Joan J. Cáceres Aug 21 '17 at 4:55
  • $\begingroup$ @JoanJ.Cáceres : No. The lemma says "if $\alpha(x)$ is nice and (some integral) holds for every function $h$ having nice endpoints, then result." This means you may assume that it works for any function you care to write down that happens to have nice endpoints. This is why the proof checks "$h(a) = h(b) = 0$ after picking an $h$, so that we know we have chosen an $h$ for which the displayed integral in the lemma statement holds. $\endgroup$ – Eric Towers Aug 21 '17 at 13:01

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