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According to the fundamental lemma of calculus of variations, if $$ \int\limits_a^b f(x)h(x) dx = 0 $$ for an arbitrary $h(x)$, then $f(x) = 0$, for $x$ in $[a, b]$. Of course, accompanied by the usual assumptions.

Now, for the case where $$ f_1(b)h(b) + \int\limits_a^b f_2(x)h(x) dx = 0 $$ again for an arbitrary $h(x)$, can we say anything similar about $f_1(b)$ and $f_2(x)$, for $x$ in $[a, b]$?

jsp13
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1 Answers1

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My version of the "fundamental lemma of calculus of variation" (e.g. from Gelfand and Fomin's book) is the following:

If $f$ is continuous on $[a,b]$ and if $\int_a^b f(x) h(x) dx=0$ for all $h$ continuous on $[a,b]$ such that $h(a)=h(b)=0$, then $f(x)=0$ for all $x\in[a,b]$.

Therefore, if you assume likewise that $f_2$ is continuous, then the condition of the lemmas is satisfied (because you can assume $h(b)=0$), so $f_2(x)=0$ for all $x\in[a,b]$. If your assumption is $h$ is any continuous function (without restricting $h(a)=h(b)=0$), then you can also deduce that $f_1(b)=0$.

md5
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  • I'm interested in the case where $h$ is any continuous function in $[a,b]$. What can we say for that case? – jsp13 May 09 '20 at 21:42
  • @jsp13: you can get $f_2(x)=0$ for all $x\in[a,b]$ (apply the lemma) and $f_1(b)=0$ (pick $h$ to be the constantly 1 function) – md5 May 10 '20 at 01:31
  • But how do you apply the lemma independently for the integral part of the left hand side? – jsp13 May 10 '20 at 13:11
  • @jsp13: Sorry but I don't understand your question. If $f_1(b)h(b)+\int_a^b f_2(x)h(x)dx=0$ for all continuous $h$, then $\int_a^b f_2(x) g(x) dx=0$ for all continuous $g$ such that $g(a)=g(b)=0$ – md5 May 10 '20 at 13:37
  • Thanks. I'm just asking how you prove this last statement of yours. – jsp13 May 11 '20 at 15:40
  • @jsp13: You mean the one in my last comment? Suppose that $f_1(b)h(b)+\int_a^b f_2(x)h(x) dx=0$ for all continuous $h$. Let $g$ be continuous such that $g(a)=g(b)=0$. We apply our hypothesis to $g$: $f_1(b)g(b)+\int_a^b f_2(x)g(x)dx=0$ and $g(b)=0$ so $\int_a^b f_2(x)g(x) dx=0$. – md5 May 11 '20 at 15:49
  • That's what I meant. Great, thanks! – jsp13 May 11 '20 at 15:53