Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2009 / A / 3

The Baire category theorem says that a nonempty complete metric space is not a countable union of closed sets with empty interior; equivalently, every countable intersection of open dense sets is dense. To prove the latter, start with any nonempty open set $V$ and open dense sets $G_n$. Choose closed balls of positive radii withEach choice is possible by density and openness. Their centers form a Cauchy sequence; its limit exists by completeness and belongs to every closed ball, hence to $V\cap \bigcap G_n$. This proves density and the theorem.

The Uniform boundedness principle says that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Z$ has uniformly bounded operator norms. Put $E_m=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le m\}$. These sets are closed and cover $X$. Baire gives an open ball $B(x_0,r)\subset E_m$. For $\Vert h\Vert <r$, the triangle inequality applied to $T(x_0+h)-T{x}_0$ gives $\Vert Th\Vert \le 2m$. Taking $h=(r/2)u$ for unit $u$ proves ${\sup }_T\Vert T\Vert \le 4m/r$.

For the particular expansion, uniqueness makes all coefficient maps and partial-sum maps linear. Since $P_{k}x\to x$, the norms $\Vert P_{k}x\Vert$ form a bounded sequence with limit $\Vert x\Vert$. Also, with $P_0=0$,This proves both preliminary bounds. The supremum therefore defines a finite norm $\Vert \cdot {\Vert }^{\prime }$, with $\Vert x\Vert \le \Vert x{\Vert }^{\prime }$.

To prove (i) implies (ii), the identity map from $(X,\Vert \cdot {\Vert }^{\prime })$ to $(X,\Vert \cdot \Vert )$ is a bounded linear bijection. If (i) holds both spaces are Banach, so the bounded inverse theorem makes its inverse bounded: $\Vert x{\Vert }^{\prime }\le C\Vert x\Vert$. This is precisely (ii).

(ii) implies (i): together with the preliminary estimate, (ii) gives equivalent norms,An equivalent norm on a Banach space is complete: a primed-norm Cauchy sequence has an original-norm limit, and the upper bound makes convergence hold also in the primed norm. (ii) implies (iii) immediately because $\Vert P_k\Vert \le C$ for every $k$.

(iii) implies (ii): the preliminary bounds show ${\sup }_k\Vert P_{k}x\Vert <\infty$ for every $x$. Applying the Uniform boundedness principle to the bounded maps $P_k$ gives ${\sup }_k\Vert P_k\Vert <\infty$.

(iii) implies (iv): the normalization $\Vert x_n\Vert =1$ givesHence every coordinate functional is bounded.

(iv) implies (iii): $P_k={\sum }_{n=1}^k{x}_n{\phi }_n$ is a finite sum of bounded rank-one operators.

(iv) implies (v): continuity gives ${\phi }_n(y_r)\to 0$ whenever $y_r\to 0$, so any scalar limit $\beta$ must vanish.

(iv) implies (vi): fix $n$. Continuity gives ${\phi }_n(y_r)\to {\phi }_n(x)$, while the assumed uniform coordinate convergence gives ${\phi }_n(y_r)\to {\beta }_n$. Uniqueness of the scalar limit yields ${\beta }_n={\phi }_n(x)$ for every $n$.

(v) implies (iv). Fix $n$ and consider any sequence $y_r\to y$ with ${\phi }_n(y_r)\to \gamma$. Apply (v) to $y_r-y$ and $\beta =\gamma -{\phi }_n(y)$. It gives $\gamma ={\phi }_n(y)$. Thus the graph of ${\phi }_n:X\to \mathbb{K}$ is closed. Both spaces are Banach, so the closed graph theorem makes ${\phi }_n$ bounded.

(vi) implies (iv) without assuming beforehand that any coordinate functional is continuous. Define the everywhere-defined linear mapThe preliminary coordinate bound ensures it really takes values in ${\ell }^{\infty }$. Condition (vi) states exactly that its graph is sequentially closed: $y_r\to x$ and $\Phi (y_r)\to \beta$ in the supremum norm imply $\Phi (x)=\beta$. The product is a metric space, so sequentially closed means closed. The closed graph theorem therefore makes $\Phi$ bounded. Every coordinate projection on ${\ell }^{\infty }$ is bounded, hence so is each ${\phi }_n$.

Together with the preceding implications this proves all six conditions are equivalent. No continuity of the coordinate functionals was assumed during the preliminary bounds.