An integral constraint and its constant multiplier

Given: The Axiom of Choice and a real number $A\mathrm{\ne }0$, the space $H_0^1(0,1)$ with its weak derivative $D$ and norm (The notation $H^k$ and the reserved zero-boundary symbol, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure), the functional $J(u)={\int }_0^1{u}^{\mathrm{\prime }2}dx$ and the constraint $G(u)={\int }_0^{1}udx-A$.

The Axiom of Choice, AC implies DC implies countable choice: the assumed Axiom of Choice supplies Dependent and Countable Choice for the integration and trace suppliers.

$H^k$ is a Hilbert space under the derivative-sum inner product, A closed subspace of a Banach space is Banach, Zero-boundary Sobolev space as a norm closure, Endpoint trace commutes with Sobolev truncation on an interval, One-dimensional $W^{1,p}$ functions have unique absolutely continuous representatives: for $I=(0,1)$ and $p=2$, the endpoint trace $T$ is well defined and $\ker T=W_0^{1,2}(I)=H_0^1(I)$; the closure defining $H_0^1(I)$ is a closed linear subspace of the real Hilbert space $H^1(I)$, hence complete for the restricted derivative-sum inner product and itself a real Hilbert space; a class in $H_0^1(I)$ has an absolutely continuous representative $v^{\ast }$ on $[0,1]$ with $v^{\ast }=0$ at both endpoints and with $(v^{\ast }{)}^{\mathrm{\prime }}=Dv$ almost everywhere.

Classical derivatives agree with weak derivatives: a $C^1$ function on $I$ has its classical derivative as weak derivative; in particular $u_0$ and the affine function $1-2x$ are weakly differentiable with $u_0^{\mathrm{\prime }}(x)=6A(1-2x)$ and $(1-2x{)}^{\mathrm{\prime }}=-2$.

The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$, For a natural $n\ge 1$ the function $x\mapsto x^n$ is differentiable everywhere with derivative $\iota (n)x^{n-1}$; for $n=0$ it is the constant $1$, with derivative $0$; for a natural $n\ge 1$ the function $x\mapsto x^{-n}$ is differentiable at every $x\mathrm{\ne }0$ with derivative $-\iota (n)x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$, The Lebesgue integral is linear on $L^1(\mu )$: the fundamental theorem of calculus applied to the antiderivatives $x^2\mathrm{/}2$ and $x^3\mathrm{/}3$, whose derivatives are $x$ and $x^2$, gives ${\int }_0^{1}xdx=1\mathrm{/}2$ and ${\int }_0^1{x}^{2}dx=1\mathrm{/}3$, and the integral is linear, so ${\int }_0^1{u}_{0}dx=6A(1\mathrm{/}2-1\mathrm{/}3)=A$.

Integration by parts for absolutely continuous functions: for absolutely continuous $F,H$ on $[0,1]$, ${\int }_0^{1}F{H}^{\mathrm{\prime }}+{\int }_0^1{F}^{\mathrm{\prime }}H=F(1)H(1)-F(0)H(0)$.

Fréchet derivative between Banach spaces, Holder's inequality for integrals, including the endpoint cases: for $u,h\in H_0^1(0,1)$ one has $J(u+h)-J(u)=2{\int }_0^1{u}^{\mathrm{\prime }}{h}^{\mathrm{\prime }}+{\int }_0^1(h^{\mathrm{\prime }}{)}^2$, and $\mid {\int }_0^1(h^{\mathrm{\prime }}{)}^2\mid =\mathrm{\parallel }{h}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }h{\mathrm{\parallel }}_{H_0^1}^2$; hence $DJ(u)h=2{\int }_0^1{u}^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$ with $\mathrm{\parallel }DJ(u)\mathrm{\parallel }\le 2\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_2$. Similarly $G$ is continuous affine, with bounded linear derivative $DG(u)h={\int }_0^{1}h$ for every $u$, and $\mathrm{\mid }DG(u)h\mathrm{\mid }\le \mathrm{\parallel }h{\mathrm{\parallel }}_2\le \mathrm{\parallel }h{\mathrm{\parallel }}_{H_0^1}$, so $G$ is $C^1$ with this derivative at every point.

The Lagrange multiplier rule for one regular constraint in Hilbert space: if $u$ is a local minimiser of $J$ on the level set $\{G=G(u)\}$ and $DG(u)\mathrm{\ne }0$, then there is a unique $\lambda \in \mathbb{R}$ with $DJ(u)=\lambda DG(u)$.

Fundamental theorem of calculus for absolutely continuous functions: an absolutely continuous function whose derivative vanishes almost everywhere is constant.