Past exam of the analysis course of the University of Oxford / 2008 / B / 5 / Solution

Use the convention that the inner product is linear in its first argument. If $v\ne 0$, nonnegativity of $\Vert u-⟨u,v⟩v/\Vert v{\Vert }^2{\Vert }^2$ givesThe case $v=0$ is immediate. This Cauchy-Schwarz inequality shows that $f_v$ is bounded linear with norm at most $\Vert v\Vert$. For $v\ne 0$, evaluation at $v/\Vert v\Vert$ gives equality; for $v=0$ both norms vanish. The Banach dual is the space of all bounded linear functionals with the operator norm.

The Riesz representation theorem asserts that every $f\in V^{\prime }$ is uniquely $f(u)=⟨u,v⟩$ for some $v\in V$, with $\Vert f\Vert =\Vert v\Vert$. If $f=0$, take $v=0$. Otherwise put $W=\ker f$, a closed subspace, and use the projection theorem to obtain $w\in W^{\perp }$ with $f(w)=1$. Since $u-f(u)w\in W$, $⟨u,w⟩=f(u)\Vert w{\Vert }^2$. Thus $v=w/\Vert w{\Vert }^2$ represents $f$. If two vectors represent it, their difference has inner product zero with every vector, in particular itself, giving uniqueness. The norm statement was already proved.

The space $P_n$ is finite-dimensional and hence complete; evaluation at zero is bounded there by equivalence of finite-dimensional norms. Riesz therefore supplies exactly one $q_n\in P_n$. In any orthonormal polynomial basis,For the real Legendre functions the bars may be omitted.

The printed even-degree normalization has an error. Orthogonality forces a degree-two polynomial to be a scalar multiple of $3t^2-1$, whose squared integral on $[-1,1]$ is $8/5$. Hence any orthonormal choice has $|p_2(0){|}^2=5/8$, whereas the printed formula gives $3/8$. The factor under its square root must be $4m+1$ instead of $2m+1$. Signs of the individual polynomials can be chosen so that the values at zero are positive; these signs do not affect $q_n$ or its norm.

Put $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$. With the consistent orthonormal normalization the norm isThe finite sum telescopes because $c_m=(2m-1)c_{m-1}/(2m)$, so $(2m+1{)}^2{c}_m^2-(2m-1{)}^2{c}_{m-1}^2=(4m+1)c_m^2$. If one substitutes the erroneous printed values formally instead, one obtains $\frac{1}{2}{\sum }_{m=0}^M(2m+1)c_m^2$; that is not the norm of the actual orthonormal Legendre evaluation representer.

For a polynomial $r$ of degree $d$, the defining identity gives $⟨r,q_n⟩=r(0)$ for every $n\ge d$, so the requested limit is eventually constant. It does not extend to all of $L^2$. First, evaluation at a point is not well-defined on equivalence classes. More strongly, the functional norms $\Vert q_n\Vert$ are unbounded. Indeed the largest binomial coefficient in row $2m$ is at least the average, so $c_m\ge 1/(2m+1)$, and the displayed sum dominates a divergent constant multiple of the harmonic series. The Uniform boundedness principle then provides an $L^2$ function for which ${\sup }_n|⟨r,q_n⟩|=\infty$. Thus no choice of point representative can restore the assertion for all $L^2$ functions.