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The geometry of a strictly convex normed space proves uniqueness whenever a best approximation exists. If distinct $p_1,p_2$ both have best error $E>0$, then their two residuals have norm $E$. The defining midpoint inequality givescontradicting best approximation because the midpoint lies in $A$. If $E=0$, both approximants equal $u$.

The printed existence assertion is false for an arbitrary vector subspace. For a concrete counterexample take $B={\ell }^2$ with its strictly convex Hilbert norm, let $A=c_{00}$ be its finite-support sequences, and take $u=(2^{-j}{)}_{j\ge 1}$. Truncations in $A$ tend to $u$, so the best-error infimum is zero, but $u\notin A$ and that infimum is not attained. If the finite-dimensional assumption of part (b) is retained, it supplies existence and the argument above supplies uniqueness.

For the estimate, suppose $Y$ is a projection operator onto $A$, as needed for a finite operator norm. Since $Yp=p$,The triangle inequality and operator norm giveThe estimate actually holds without attainment: apply it with any $q\in A$ instead of $p$ and take the infimum. Its proof applies in every normed space. This is strict convexity separates uniqueness from existence in best approximation.

Let $V_h$ be the continuous piecewise linear space including both endpoint hat functions. Robin conditions are natural conditions and do not restrict the endpoint values of this space. The finite element problem is $u_h\in V_h$, $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. coercivity restricted to $V_h$ gives existence and uniqueness.

Expanding $u_h={\sum }_{j=0}^N{U}_j{\varphi }_j$ gives $\mathsf{A}U=F$, whereThe stiffness contribution has diagonal $2/h$ at interior nodes, $1/h$ at endpoints, and adjacent entries $-1/h$. For completeness, the weighted mass contribution has interior diagonal $2h{x}_i^2/3+h^3/15$, left endpoint diagonal $h{a}^2/3+a{h}^2/6+h^3/30$, right endpoint diagonal $h{b}^2/3-b{h}^2/6+h^3/30$, and adjacent entry $h{x}_i^2/6+h^2{x}_i/6+h^3/20$. All more distant entries vanish. The Robin contribution adds one to each endpoint diagonal.

Symmetry follows from symmetry of the bilinear form. If $V\ne 0$, nodal independence makes $v_h={\sum }_i{V}_i{\varphi }_i$ nonzero, and $V^T\mathsf{A}V=a(v_h,v_h)\ge \gamma \Vert v_h{\Vert }_{H^1}^2>0$. Thus the matrix is symmetric positive definite.

The Sobolev space and norm areIn one dimension these functions have continuous representatives and well-defined endpoint traces. integration by parts gives the weak problem: find $u\in H^1(a,b)$ such that $a(u,v)=\ell (v)$ for all $v\in H^1(a,b)$, whereIndeed the integration boundary term is $-u^{\prime }(b)v(b)+u^{\prime }(a)v(a)$; substituting $u^{\prime }(a)=u(a)-A$ and $u^{\prime }(b)=B-u(b)$ gives the two positive boundary terms on the left.

If $0\notin [a,b]$, put $m={\min }_{[a,b]}{x}^2>0$. Then $a(v,v)\ge \Vert v^{\prime }{\Vert }_2^2+m\Vert v{\Vert }_2^2\ge \min (1,m)\Vert v{\Vert }_{H^1}^2$. If $0\in [a,b]$, the supplied trace inequality givesThus the form is coercive in both cases. The trace theorem and boundedness of $x^2$ give continuity of $a$ and $\ell$, and the Lax-Milgram theorem gives existence and uniqueness.

A strictly convex norm permits at most one best approximation from a linear subspace, but does not force the distance infimum to be attained. In the Hilbert space ${\ell }^2$, the finite-support vector subspace $c_{00}$ is dense and omits $u=(2^{-j}{)}_{j\ge 1}$. Its distance from $u$ is zero and no member attains that distance. Finite-dimensionality restores existence by compactness of bounded sets in the vector subspace. If $Y$ is a bounded projection onto a vector subspace, the estimate $\Vert u-Yu\Vert \le (1+\Vert Y\Vert ){\inf }_q\Vert u-q\Vert$ follows directly by writing $u-Yu=(I-Y)(u-q)$ for each candidate $q$, even when the infimum is not attained.

A solution is unique when any two objects satisfying the stated conditions are equal. This is an at-most-one assertion; existence is a separate requirement. A common proof assumes two solutions, subtracts their equations, and proves that their difference vanishes. For a linear differential equation this turns the task into a homogeneous problem with zero boundary or initial data.