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this is a proof of the fundamental lemma of calculus of variation. Some preparations: Let $g(x):=e^{\frac{-1}{1-||x||}} \chi_{||x||<1},$ with characteristic function $\chi,$ then $$c:= \int_{\mathbb{R}^n}g(||x||^2)dx$$ and $\delta_k(x):=\frac{k^n}{c} g(||kx||^2)$ with $||\delta_k||_{L^1} = 1$ and $supp(\delta_k) = B_{[0,\frac{1}{k}]}(0).$ So we have that $f \in L^1$ has the property that $\delta_k * f \rightarrow f$ in $L^1$. This is basically how you show that $C^{\infty}$ is dense in $L^1.$

The fundamental lemma of the calculus of variation says now that if for $f \in L^1_{loc}(\Omega)$ we have $\int_{\Omega} f\phi dx = 0$ for all testfunctions $\phi$, then we already get $f=0.$

First I assume that $\Omega$ is bounded, because the general case is an easy consequence of this. Now define $\Omega_{\delta}:=\{x \in \Omega: d(x,\partial \Omega) \ge \delta\}$ and $f_m:= f|_{\Omega_m}.$

Now, $f_m * \delta_k \in C_c^{\infty}(\Omega)$ for $k$ large enough and $f_m * \delta_k \rightarrow f_m$ in the $L^1$ sense.

At this point, I know that $0 = \int_{\Omega} f(\delta_k * f_m)$ by the assumption of the fundamental lemma.

I definitely get a subsequence $(\delta_{k_j} * f_m)$ that converges pointwise a.e. to $f_m$ and I want to apply Lebesgue's theorem to conclude

$0 = \int_{\Omega} f(\delta_{k_j} * f_m) \rightarrow \int_{\Omega_m} |f_m|^2$ which shows that $f_m = 0$ a.e. for all m , so $f=0$ a.e., but I don't see why the condition of the dominated convergence theorem is fulfilled here.

Please do not post different proofs to this Lemma, I want to understand this last piece of this one.

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  • $\begingroup$ This isn't a homework problem, is it? $\endgroup$
    – Nik Weaver
    Apr 4, 2015 at 23:25
  • $\begingroup$ no, in a homework problem I would ask: Hey, I am too lazy to do it, please give me a proof of theorem XYZ. No honestly, I am actually interested how the dominated convergence theorem could be applied here or if there is no way out? $\endgroup$
    – Physicist 2.0
    Apr 4, 2015 at 23:29
  • $\begingroup$ But the fix is easy: instead of using $\delta_{k_j} * f_m$, use $\delta_{k_j} * \operatorname{sgn}(f_m)$. Then we get $|\delta_{k_j} * \operatorname{sgn}(f_m)| \le 1$ so the integrand is dominated by $|f|$ which is integrable over $\Omega$. And the limit is $\int |f_m|$. $\endgroup$ Apr 5, 2015 at 1:43
  • $\begingroup$ @NateEldredge thank you. well, I basically have the proof from here and they also give a reason why you can apply Lebesgue's theorem, but I don't see what they are doing there and I assume it to be wrong (see page 38 on this pdf file math.tu-berlin.de/fileadmin/i26_fg-kutyniok/Fritz/FA2_WS1415/…) $\endgroup$
    – Physicist 2.0
    Apr 5, 2015 at 1:48
  • $\begingroup$ Yes, I believe their proof is incorrect. $\endgroup$ Apr 5, 2015 at 3:34

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(I write this as an answer because it is a bit longer than a comment.)

The problem is with the statement $\delta_k\ast f\to f$ in $L^1$. Why does this happen? You cannot invoke the dominated convergence theorem here.

For example, in Brezis' Functional analysis book this fact is proved using fundamental lemma of variational calculus. Other proofs use the fact that $C_{cpt}(\mathbb{R}^n)$ is dense in $L^1(\mathbb{R}^n)$. This follows from Urysohn's theorem on the abundance of continuous function. This density factn cannot be proved using convolutions, and it is also responsible for the fundamental lemma. Like Nate Eldredge, I believe that the proof of the fundamental lemma you outlined is incorrect.

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