Past exam of the analysis course of the University of Oxford / 2010 / A / 2

Equip the continuous dual space $X^{\prime }$ with the operator normContinuity of a linear functional makes this finite; homogeneity, definiteness and the triangle inequality follow directly from the supremum formula.

For completeness of the dual space, let $(f_n)$ be Cauchy in this norm. For each $x\in X$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $(f_n(x))$ is Cauchy in $\mathbb{R}$. Define $f(x)={\lim }_n{f}_n(x)$. Passing to the limit in the linearity identities makes $f$ linear. Since a Cauchy sequence is bounded, ${\sup }_n\Vert f_n\Vert \le C$ and $|f(x)|\le C\Vert x\Vert$, so $f\in X^{\prime }$.

Given $\epsilon >0$, choose $N$ with $\Vert f_n-f_m\Vert \le \epsilon$ for $n,m\ge N$. Let $m\to \infty$ pointwise; then $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ and therefore $\Vert f_n-f\Vert \le \epsilon$. Thus $X^{\prime }$ is a Banach space. This proof in fact works even if $X$ itself is not complete.

If a bilinear map is continuous, continuity at $(0,0)$ gives $\delta >0$ such that $\Vert F(x,y)\Vert \le 1$ whenever $\Vert x\Vert ,\Vert y\Vert <\delta$. For nonzero $x,y$, apply this to $\delta x/(2\Vert x\Vert )$ and $\delta y/(2\Vert y\Vert )$. Bilinearity givesIf either input is zero, the inequality follows from bilinearity directly.

Conversely, suppose the product bound holds with $C\ge 0$. At a fixed point $(x,y)$,Its norm is at most $C(\Vert h\Vert \Vert y\Vert +\Vert x\Vert \Vert k\Vert +\Vert h\Vert \Vert k\Vert )$, which tends to zero as $(h,k)\to (0,0)$. Thus the product bound is equivalent to joint continuity. This is boundedness criterion for a bilinear map.

Put $\rho =r-\sqrt{r^2-1}$. Since $r>1$, $0<\rho <1$ and ${\rho }^2-2r\rho +1=0$. The nonzero square-summable sequence ${\beta }_j={\rho }^{j-1}$ defines a continuous linear functional on ${\ell }_2$ byThe Cauchy-Schwarz inequality proves continuity. For every $k$,Hence $f$ vanishes on the closed linear span $M$, but $f(e_1)=1$. ThereforeMore precisely, any vector in the orthogonal complement of $M$ satisfies the recurrence ${\beta }_{k+2}-2r{\beta }_{k+1}+{\beta }_k=0$. Its roots are $r\pm \sqrt{r^2-1}$, one less than one and the other greater than one. Square summability eliminates the growing solution, so $M^{\perp }$ is exactly the span of $({\rho }^{j-1})$. Thus $M=\ker f$ is a closed hyperplane of codimension one. This is annihilator of a second-difference span in l2.