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The one-dimensional Sobolev representative of $u$ is absolutely continuous, and its zero trace givesConsequently $u(x)/x=A(u^{\prime })(x)$ for the Hardy operator. Applying the Hardy averaging inequality to $u^{\prime }$ gives the Hardy inequality on an interval:

Multiply the differential equation by $v\in H$ and integrate. The Hardy inequality on an interval makes the terms containing $u/x$ and $v/x$ integrable. For smooth $v$, integration by parts givesbecause $v(0)=0$ and the Neumann boundary condition is $u^{\prime }(1)=0$. ThereforeThe density of smooth functions in a Sobolev space and continuity of all four terms extend this weak formulation to every $v\in H$.

For $u\in H$, the trace $u(0)$ vanishes. Apply the Hardy inequality on an interval with $p=2$ to obtainThus $u/x\in L^2(0,1)$.

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincare inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.

On $H=\{v\in H^1(0,1):v(0)=0\}$, the formis bounded and coercive. The Hardy inequality on an interval controls its singular term, while the one-sided Poincare inequality controls the first-order term. Thus the Lax-Milgram theorem gives a unique weak solution for every functional $v\mapsto {\int }_0^1(f/x)(v/x)$ with $f/x\in L^2(0,1)$.