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What I want to know is how do you prove the fundamental lemma of the calculus of variations without knowing what a ball is. I saw a proof but you needed to know that you could substitute "characteristics of balls" for "smooth compactly supported functions", and I didn't understand why you could exchange phrases. I think I know how to prove this, but I'm not sure. Here's the lemma:

If $f(x)$ is continuous on the open interval $(a, b)$ and for all compactly supported smooth functions $h(x)$ on $(a,b)$,

$$\int_a^b f(x)h(x) =0$$

then $f(x)$ is identically zero.

If the hypothesis entails the assumption that $h(x)$ is integrable, then I can prove it myself, but I don't know if infinitely differentiable implies integrable. I understand that differentiability implies continuity, and I think if a function is continuous on $[a,b]$ then it's integrable on the $closed$ interval $[a,b]$, but if the interval is open I'm not so sure. Any help would be greatly appreciated.

Here's my proof. Assume $h(x)$ can be integrated on $(a,b)$, then $h(x)=v'$. So we are interested in $\int_a^b f(x) dv=0$. Using integration by parts we have:

$\int f(x)dv =f(x)v - \int v(x) df=0$

Since h(x) can be any smooth function take $v(x)=x$. Then

$0 =f(x)x - \int x df=f(x)x -f(x)x - \int_a^b f(x)dx$

Which is true only if $f(x)=0.$

lee pappas
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    infinitely-diffeentiable implies continuous, and continuous functions on $[a,b]$ are integrable. The standard proof uses uniform continuity… google it – peek-a-boo Mar 16 '24 at 23:26
  • @peek-a-boo that is not really a helpful comment since if the function is only defined on $(a,b)$, it may tend to infinity near $a$ or $b$, or do something else, making it not continuously extendable to $[a,b]$, or is there something I'm missing? – Neckverse Herdman Mar 17 '24 at 00:21
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    An edit to my comment: I think $h$ is indeed still integrable because it is compactly supported in $(a,b)$, so can find $a<c<d<b$ such that $h$ vanishes outside $[c,d]$ and on $[c,d]$ it is integrable by continuity and $[c,d]$ being closed and bounded – Neckverse Herdman Mar 17 '24 at 00:32
  • Thank you both. I have added my proof, is it OK? – lee pappas Mar 17 '24 at 01:24
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    @NeckverseHerdman the question was changed after I made my comment. Anyway, like you said in your subsequent comment, compact support in $(a,b)$ makes the question all the more trivial. – peek-a-boo Mar 17 '24 at 01:35

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