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

The quotient norm isChanging the representative does not change the infimum. The triangle inequality and absolute homogeneity follow from those of the original norm. Vanishing means $x$ belongs to the closure of $Y$, hence to $Y$ because the subspace is closed. Thus this is a norm.

Use the absolutely convergent series criterion for completeness. If $\sum \Vert {\xi }_n{\Vert }_{X/Y}<\infty$, choose representatives $z_n$ of ${\xi }_n$ satisfying $\Vert z_n\Vert \le \Vert {\xi }_n{\Vert }_{X/Y}+2^{-n}$. The series $\sum z_n$ converges in the Banach space $X$ to some $z$. Since the quotient map is contractive, $\sum {\xi }_n$ converges to $z+Y$. Consequently the quotient is a Banach space.

A set is totally bounded if for every $\epsilon >0$ it can be covered by finitely many open balls of radius $\epsilon$.

Here is the needed separated-sequence argument, including the relevant Riesz lemma. For a proper closed subspace $Y$, choose $z\notin Y$, put $d=\mathrm{dist}(z,Y)>0$, and choose $y\in Y$ with $\Vert z-y\Vert <2d$. Then $w=(z-y)/\Vert z-y\Vert$ is a unit vector at distance greater than $1/2$ from $Y$. Starting with one unit vector, apply this to the span of the vectors already chosen. These spans are closed and proper because they are finite-dimensional while $X$ is infinite-dimensional. The resulting unit vectors satisfy $\Vert w_m-w_n\Vert >1/2$ for $m\ne n$. A ball of radius $1/4$ contains at most one of them. Hence the closed unit ball is not totally bounded.

Take $S=T(B_X)$. A bounded linear operator is compact exactly when the closure of this set is compact. In a complete metric space this is equivalent to its being totally bounded.

If $T$ is compact, choose a finite $\epsilon$-net $z_1,\dots ,z_m$ for $T(B_X)$ and let $Y$ be their span. Then $\mathrm{dist}(Tx,Y)<\epsilon$ for $\Vert x\Vert \le 1$, giving $\Vert {\pi }_{Y}T\Vert \le \epsilon$.

Conversely, fix $\delta >0$ and take $Y$ with $\Vert {\pi }_{Y}T\Vert \le \delta /4$. For every $x\in B_X$ choose $y_x\in Y$ with $\Vert Tx-y_x\Vert <\delta /2$. These approximants satisfy $\Vert y_x\Vert \le \Vert T\Vert +\delta /2$, so form a bounded subset of a finite-dimensional space and have a finite $\delta /2$-net. The same centers give a finite $\delta$-net for $T(B_X)$. Thus it is totally bounded and $T$ is compact. In particular

The natural canonical embedding into the bidual is $J_{X}x(f)=f(x)$. It is linear and $\Vert J_{X}x\Vert \le \Vert x\Vert$. A norming functional from the Hahn-Banach theorem gives the reverse inequality, so it is an isometry. A reflexive Banach space is one for which this canonical embedding is onto.

A nonreflexive example is $c_0$. Its dual is ${\ell }^1$ and its bidual is ${\ell }^{\infty }$; the canonical map is the natural inclusion, which misses the constant sequence one. This also shows why merely having an isometric embedding in the bidual is insufficient.

An infinite-dimensional reflexive example is ${\ell }^2$, by the Riesz representation theorem.

The weak topology on $X$ is the coarsest topology making every $f\in X^{\ast }$ continuous. The weak-star topology on $X^{\ast }$ is the coarsest making every evaluation $f\mapsto f(x)$, $x\in X$, continuous. In each case basic neighborhoods impose finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of a dual space is compact for its weak-star topology. The separation theorem needed here says: if $K$ is a nonempty weak-star compact convex subset of a dual space $E^{\ast }$ and $z\notin K$, then some $e\in E$ strictly separates them,The real-part convention covers complex spaces; for real spaces omit it.

Let $K$ be the weak-star closure of $J_X(B_X)$ inside $B_{X^{\ast \ast }}$. This is compact by Banach-Alaoglu, convex, and nonempty. If $z\in B_{X^{\ast \ast }}\setminus K$, separation in the dual pair $(X^{\ast \ast },X^{\ast })$ gives $f\in X^{\ast }$ withBut $|z(f)|\le \Vert z\Vert \Vert f\Vert \le \Vert f\Vert$, a contradiction. Therefore $K=B_{X^{\ast \ast }}$, proving the required Goldstine theorem density.

If $X$ is reflexive, $J_X$ identifies its weak topology with the weak-star topology of $X^{\ast \ast }$ and carries $B_X$ onto $B_{X^{\ast \ast }}$, so Banach-Alaoglu gives weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact, hence closed in the Hausdorff bidual topology. Density then forces $J_X(B_X)=B_{X^{\ast \ast }}$. Scaling gives surjectivity on the entire bidual. Thus

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.