The Baire Principles of Functional Analysis

Put $E_n=\{x:{\sup }_{T\in \mathcal{F}}\mathrm{\parallel }Tx\mathrm{\parallel }\le n\}$. Each $E_n$ is closed (an intersection of inverse images of closed balls), and pointwise boundedness gives $X={\bigcup }_{n\ge 1}{E}_n$.

By Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior, some $E_N$ contains a ball $B(x_0,r)$.

For $\mathrm{\parallel }h\mathrm{\parallel }<r$, both $x_0$ and $x_0+h$ lie in that ball, so $\mathrm{\parallel }Th\mathrm{\parallel }\le \mathrm{\parallel }T(x_0+h)\mathrm{\parallel }+\mathrm{\parallel }T{x}_0\mathrm{\parallel }\le 2N$ for every $T$.

Rescaling a nonzero $x$ to $h=rx\mathrm{/}(2\mathrm{\parallel }x\mathrm{\parallel })$ yields $\mathrm{\parallel }Tx\mathrm{\parallel }\le 4N\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}r$; the same is clear for $x=0$. Thus $\mathrm{\parallel }T\mathrm{\parallel }\le 4N\mathrm{/}r$ for every $T$.

Let $E_n=\{x:{\sup }_T\mathrm{\parallel }Tx\mathrm{\parallel }\le n\}$. These sets are closed, and $X\setminus S={\bigcup }_n{E}_n$.

If some $E_n$ has nonempty interior, the translation-and-rescaling argument of Uniform boundedness principle gives a common operator-norm bound.

Otherwise every $E_n$ is closed with empty interior. Its complement is open dense, so $S={\bigcap }_n(X\setminus E_n)$ is $G_{\delta }$ and dense by Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior.

The two alternatives exhaust the cases from step 2.1, proving the dichotomy.

Pointwise convergence makes $(T_{n}x{)}_n$ bounded for each $x$. Thus Uniform boundedness principle gives $M:={\sup }_n\mathrm{\parallel }{T}_n\mathrm{\parallel }<\mathrm{\infty }$.

Passing $T_n(x+y)=T_{n}x+T_{n}y$ and $T_n(\lambda x)=\lambda T_{n}x$ to limits shows that $T$ is linear.

For each $x$, continuity of the norm gives $\mathrm{\parallel }Tx\mathrm{\parallel }={\lim }_n\mathrm{\parallel }{T}_{n}x\mathrm{\parallel }\le M\mathrm{\parallel }x\mathrm{\parallel }$, so $T$ is bounded.

If ${\mathrm{lim\hspace{0.17em}inf}}_n\mathrm{\parallel }{T}_n\mathrm{\parallel }=\mathrm{\infty }$, the asserted bound is immediate. Otherwise select a subsequence whose norms tend to the finite liminf. The preceding inequality along that subsequence gives $\mathrm{\parallel }Tx\mathrm{\parallel }\le ({\mathrm{lim\hspace{0.17em}inf}}_n\mathrm{\parallel }{T}_n\mathrm{\parallel })\mathrm{\parallel }x\mathrm{\parallel }$ for every $x$, hence the asserted norm bound.

The unit-ball definition of the operator norm supplies $z$ with $\mathrm{\parallel }z\mathrm{\parallel }\le 1$ and $\mathrm{\parallel }Tz\mathrm{\parallel }>8\mathrm{\parallel }T\mathrm{\parallel }\mathrm{/}9$.

Put $w=(3r\mathrm{/}4)z$. Then both $x+w$ and $x-w$ lie in $B(x,r)$, and $\mathrm{\parallel }Tw\mathrm{\parallel }>2r\mathrm{\parallel }T\mathrm{\parallel }\mathrm{/}3$.

The triangle inequality applied to $T(x+w)-T(x-w)=2Tw$ gives $\max \{\mathrm{\parallel }T(x+w)\mathrm{\parallel },\mathrm{\parallel }T(x-w)\mathrm{\parallel }\}\ge \mathrm{\parallel }Tw\mathrm{\parallel }>\frac{2r}{3}\mathrm{\parallel }T\mathrm{\parallel }\mathrm{.}$

Choosing the corresponding one of $x+w,x-w$ proves the claim.

Suppose the norms are unbounded. Countable choice selects $T_n\in \mathcal{F}$ with $\mathrm{\parallel }{T}_n\mathrm{\parallel }\ge 4^n$ for $n\ge 1$. Set $x_0=0$.

Recursively, apply A nonzero bounded linear operator is large on one of two nearby points with centre $x_{n-1}$ and radius $3^{-n}$ to choose $x_n$ with $\mathrm{\parallel }{x}_n-x_{n-1}\mathrm{\parallel }<3^{-n}$ and $\mathrm{\parallel }{T}_n{x}_n\mathrm{\parallel }>(2\mathrm{/}3)3^{-n}\mathrm{\parallel }{T}_n\mathrm{\parallel }$. DC licenses these dependent choices.

The sequence $(x_n)$ is Cauchy, since its tails are bounded by a tail of ${\sum }_{n\ge 1}{3}^{-n}$; completeness gives $x_n\to x\in X$. Moreover $\mathrm{\parallel }x-x_n\mathrm{\parallel }\le {\sum }_{k>n}{3}^{-k}=3^{-n}\mathrm{/}2$.

Hence $\mathrm{\parallel }{T}_{n}x\mathrm{\parallel }\ge \mathrm{\parallel }{T}_n{x}_n\mathrm{\parallel }-\mathrm{\parallel }{T}_n\mathrm{\parallel }\mathrm{\parallel }x-x_n\mathrm{\parallel }>\frac{1}{6}{3}^{-n}\mathrm{\parallel }{T}_n\mathrm{\parallel }\ge \frac{1}{6}(4\mathrm{/}3{)}^n,$ which contradicts pointwise boundedness at $x$.

Surjectivity gives $Y={\bigcup }_{n\ge 1}\overline{T(B_X(0,n))}$. Baire applied to $Y$ gives $n$, $y_0$, and $\rho >0$ with $B(y_0,\rho )$ inside this closure.

Subtracting two points in this ball shows $B(0,\rho )\subseteq \overline{T(B(0,2n))}$.

Scaling by $2n$ gives $B(0,\rho \mathrm{/}(2n))\subseteq \overline{T(B(0,1))}$, as required.

Given $y$ with $\mathrm{\parallel }y\mathrm{\parallel }<r\mathrm{/}2$, repeatedly use the closure inclusion, after scaling, to choose $x_n$ with $\mathrm{\parallel }{x}_n\mathrm{\parallel }<2^{-n-1}$ and residual $y-T{\sum }_{k\le n}{x}_k$ of norm $<r{2}^{-n-2}$. DC licenses this dependent recursive selection.

The series ${\sum }_n{x}_n$ converges in $X$ by Series criterion for Banach spaces, and its sum $x$ has $\mathrm{\parallel }x\mathrm{\parallel }<1$.

Boundedness makes $T$ continuous by For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, so $Tx={\lim }_{n}T{\sum }_{k\le n}{x}_k=y$ as the residuals vanish.

The preceding successive-approximation lemma gives $\epsilon >0$ with $B_Y(0,\epsilon )\subseteq T(B_X(0,1))$.

For every $x\in X$ and $a>0$, linearity gives $B_Y(Tx,a\epsilon )\subseteq T(B_X(x,a))$.

Let $U\subseteq X$ be open and $y=Tx\in T(U)$. Choose $a>0$ with $B_X(x,a)\subseteq U$. Then step 2.1 gives $B_Y(y,a\epsilon )\subseteq T(U)$. Thus every point of $T(U)$ is interior, so $T(U)$ is open.

By Open mapping theorem, $T(B_X(0,1))$ is an open neighbourhood of $0$.

It therefore contains $B_Y(0,c)$ for some $c>0$, which is exactly the claim.

By Open mapping theorem, $T$ maps the open unit ball onto a neighbourhood of $0$; hence $B_Y(0,c)\subseteq T(B_X(0,1))$ for some $c>0$.

If $y\mathrm{\ne }0$, the point $cy\mathrm{/}(2\mathrm{\parallel }y\mathrm{\parallel })$ lies in $B_Y(0,c)$, so its unique preimage has norm $<1$. Scaling gives $\mathrm{\parallel }{T}^{-1}y\mathrm{\parallel }<2c^{-1}\mathrm{\parallel }y\mathrm{\parallel }$; this also holds at $0$.

Thus $T^{-1}$ is bounded, and it is linear because $T$ is a linear bijection.

If $T$ is bounded, $x_n\to x$ and $T{x}_n\to y$, continuity gives $y=Tx$, so the graph is closed.

Conversely, a closed graph is Banach by A closed subspace of a Banach space is Banach, since $X\times Y$ is Banach by Finite products of Banach spaces are Banach.

The first projection $G(T)\to X$ is a bounded linear bijection; its inverse is bounded by Bounded inverse theorem. Composing that inverse with the second projection makes $T$ bounded.

If the graph closure is a graph and $(0,y)$ lies in it, it must equal the graph point $(0,0)$; this proves the sequential condition.

Conversely, if $(x,y)$ and $(x,z)$ lie in the graph closure, sequences of graph points converging to them exist by metric closure (using DC to choose $1\mathrm{/}(n+1)$ approximants). Their differences give a sequence tending to $(0,y-z)$.

The sequential condition gives $y-z=0$, so the graph closure has at most one second coordinate over each first coordinate; as a closed linear subspace it is a graph.

For each $y$ in the unit ball of $Y$, $x\mapsto b(x,y)$ is bounded; for fixed $x$, separate continuity makes their values at $x$ bounded on that unit ball.

Uniform boundedness Uniform boundedness principle gives $C$ with $\mathrm{\parallel }b(x,y)\mathrm{\parallel }\le C\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$.

Thus $b$ is bounded in the sense of A bounded bilinear map between normed spaces, and the boundedness/joint-continuity equivalence For a bilinear map, boundedness is equivalent to joint continuity yields joint continuity.

The identity $I:(V,p)\to (V,q)$ is a bounded linear bijection by the assumed inequality.

Both spaces are Banach (Banach space), so Bounded inverse theorem makes $I^{-1}$ bounded: $p(v)\le C^{\mathrm{\prime }}q(v)$ for some $C^{\mathrm{\prime }}>0$.

The two inequalities are precisely equivalence of $p$ and $q$.