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I am looking at Theorem 2 from Evans PDE 2nd Edition chapter 8.4.1 about Lagrange Multipliers. Here he is trying to prove there is a real number $\lambda$ such that

$$\int_U Du\cdot Dv\;dx=\lambda\int_U g(u)v\;dx$$

where $u\in H_0^1(U)$ is the minimizer of a functional $I[\cdot]$ subject to an integral constraint $J(w)=\int_U G(w)\;dx=0$. The part I am confused about is in the case where he assumes $g(u)=0$ almost everywhere ($g=G'$ for $G$ smooth). He argues that

$$D(G(u))=g(u)Du=0$$

thus since $U$ is assumed to be smooth, bounded, and connected, $G(u)$ must be constant. Since $J(u)=0$, we must have $G(u)=0$ almost everywhere. So far so good. From here, he asserts that "as $u=0$ on $\partial U$ in the trace sense, it follows that $G(0)=0$"

This is where I am stuck. If $u$ were continuous, then it would be immediate since $u$ would have to be close to zero on some set near the boundary, so if $G(0)\neq0$ then this would violate $J(u)=0$. But we only know that it is zero on the boundary in the trace sense. My attempts to use density by $C_c^\infty(U)$ functions have not worked, since there is no guarantee that the approximating functions stay near to zero at the boundary in any uniform way to break the condition $J(u)=0$ in the same way.

It seems this should be rather straightforward since Evans doesn't elaborate on this detail but I seem to be missing something obvious. Any ideas?

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1 Answer 1

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If I understand correctly, the question is:

Let $U$ be a smooth bounded connected open subset of $\mathbb{R}^n$, $G : \mathbb{R} \to \mathbb{R}$ a smooth function and $u \in H^1_0(U;\mathbb{R})$ such that $G(u) = 0$ a.e. in $U$. Prove that $G(0) = 0$.

I would use the following argument:

By contradiction, if $G(0) \neq 0$, since $G$ is continuous, there exists $\varepsilon > 0$ and $\delta > 0$ such that, $|G(s)| \geq \varepsilon$ for all $s \in [-\delta,\delta]$. Since $G(u) = 0$ a.e. in $U$, the Lebesgue measure of the set $\{ x \in U ; |G(u(x))| \geq \varepsilon \}$ is zero. Thus, the Lebesgue measure of the set $\{ x \in U ; |u(x)| \leq \delta \}$ is zero. Hence $|u| \geq \delta$ a.e. in $U$. Since $u \in H^1_0(U;\mathbb{R})$, then so is $v := |u|$. Now we have a function $v \in H^1_0(U;\mathbb{R})$ which satisfies $v \geq \delta$ a.e. in $U$. From trace theory, this contradicts the fact that $v = 0$ on $\partial\Omega$. Indeed, you should know that a function which is non-negative a.e. in $U$ and in $H^1$ has an a.e. non-negative trace, which you can apply to $v-\delta \in H^1(U;\mathbb{R})$.

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