Section 4.4 Three fundamental results on mappings
We borrow the Baire Category Theorem from topology to show the uniform boundedness principle for linear operators.Theorem 4.4.1. Baire category theorem.
If \(U_i\) is a countable family of dense open sets of a complete metric space \(X\text{,}\) then,
\begin{equation*}
\bigcap_{n=1}^\infty U_n
\end{equation*}
is dense in X. Proof.
\begin{equation*}
\overline{B_{x_1,r_1}}\subset W\cap U_1.
\end{equation*}
By repeating this argument, we build sequences \(x_n,r_n\) such that
\begin{equation*}
\overline{B_{x_n,r_n}}\subset B_{x_{n-1},r_{n-1}}\cap U_n.
\end{equation*}
We can assume without loss of generality that \(r_n<1/n\text{.}\) The sequence \(x_n\) is Cauchy since \(x_n\in B_{x_m,r_m}\) for every \(n>m\text{,}\) so it converges to some \(x\in X\text{.}\) Necessarily, \(x\in \overline{B_{x_n,r_n}}\) for every \(n\) since these sets are all one inside the other. Hence, \(x\in W\cap U_1\cap U_2\cap\dots.\)Example 4.4.2.
\begin{equation*}
U_n = [0,2\pi)\backslash \left\{\frac{1}{n}\right\}.
\end{equation*}
Then
\begin{equation*}
\bigcap_n U_n = [0,2\pi)\backslash \left\{1,\frac{1}{2}, \frac{1}{3}, \dots \right\},
\end{equation*}
which is still dense in \(\bS^1\text{.}\)Corollary 4.4.3.
\(F\) is called nowhere dense in \(X\) (of the first category) if \(\overline{F}\) contains no nonempty open sets. Let \(F_j\) be a countable sequences of nowhere dense sets. Then
\begin{equation*}
\bigcup_{j=1}^{\infty} F_j \neq X.
\end{equation*}
Proof.
\begin{equation}
\bigcap U_j
=\bigcap X\setminus \overline{F}_j
=X\setminus \bigcup \overline{F}_j.\tag{4.4.1}
\end{equation}
Hence \(\bigcup \overline{F}_j \neq X\text{.}\)Theorem 4.4.4. The Uniform Boundedness Principle.
Let \(X\) be a Banach space and \(Y\) a normed space and let \(\cS\subset \cB(X,Y)\) be a family of bounded operators. Then, if \(\cS\) is pointwise bounded, it is equicontinuous. Namely, if
\begin{equation*}
\sup_{T\in \cS}\|T x\|_Y<\infty \quad \mbox{for all }\quad x\in X,
\end{equation*}
then there is a \(C>0\) such that
\begin{equation*}
\|Tx\|<C\|x\|\text{ for all } x\in X\text{ and }T\in\cS
\end{equation*}
or, in other words,
\begin{equation*}
\sup_{T\in \cS}\|T\|_{op}<\infty.
\end{equation*}
Proof.
\begin{equation*}
B_{y,r}\subset F_n.
\end{equation*}
This means that \(y+x\in F_n \) for all \(\|x\|_X\leq r\text{,}\) so that
\begin{equation}
\|Tx\|_Y=\|T(y+x)+T(-y)\|
\leq n+ \|Ty\|_Y
\leq R\tag{4.4.2}
\end{equation}
for some \(R>0\text{.}\) Since all maps in \(\cS\) are linear, we can exploit the bound above as follows. Let \(x\in X\text{.}\) Then
\begin{equation*}
\|Tx\| = \frac{\|x\|}{r}\left\|T\left(r\frac{x}{\|x\|}\right)\right\| \leq \frac{\|x\|}{r} \frac{R}{r} \left\|r\frac{x}{\|x\|}\right\| =\frac{R}{r}\|x\|.
\end{equation*}
Hence, we conclude that
\begin{equation*}
\|T\|_{op}\leq \frac{R}{r}
\end{equation*}
for all \(T\in\cS\text{.}\)Theorem 4.4.5. The Open Mapping Theorem.
Let \(X,Y\) be Banach spaces and let \(L\in\cB(X,Y)\) be onto, i.e. \(L(X)=Y\text{.}\) Then \(L\) is an open map, i.e. \(L\) takes open sets to open sets.Proof.
\begin{equation*}
B_{y,r'}\subset L(B_{x,r}).
\end{equation*}
By linearity, it is enough to prove that there is a \(c>0\) such that
\begin{equation*}
B_{0_Y,c}\subset L(B_{0_X,1}).
\end{equation*}
Notice first that
\begin{equation*}
\bigcup_{n=1}^\infty B_{0_X,n} = \bigcup_{n=1}^\infty nB_{0_X,1} = X,
\end{equation*}
so that, since \(L\) is surjective,
\begin{equation*}
\bigcup_{n=1}^\infty n\overline{L(B_{0_X,1})} = \bigcup_{n=1}^\infty \overline{L(nB_{0_X,1})} = L(X) = Y.
\end{equation*}
Since \(Y\) is complete, by the Baire theorem, there must be a \(n_0\geq1\) such that \(n_0\overline{L(B_{0_X,1})}\text{,}\) and so \(\overline{L(B_{0_X,1})}\text{,}\) contains an open ball. So, let \(c>0\) and \(y_0\in Y\) such that
\begin{equation*}
B_{y_0,c}\subset\overline{L(B_{0_X,1})}.
\end{equation*}
By symmetry, \(-y_0\in \overline{L(B_{0_X,1})}\text{,}\) so that
\begin{equation*}
B_{y_0,4c}\subset\overline{L(B_{0_X,1})}+\overline{L(B_{0_X,1})}.
\end{equation*}
By the triangle inequality, balls in a Banach space are convex and so
\begin{equation*}
\overline{L(B_{0_X,1})}+\overline{L(B_{0_X,1})}\subset\overline{L(B_{0_X,2})}.
\end{equation*}
Hence, by linearity,
\begin{equation*}
B_{0_Y,2c}\subset\overline{L(B_{0_X,1})}
\end{equation*}
or, equivalently,
\begin{equation*}
B_{0_Y,c}\subset\overline{L(B_{0_X,\frac{1}{2}})}
\end{equation*}
This means that, for every \(\eps>0\) and \(y\in Y\) with \(\|y\|< c\text{,}\) there is a \(x\) with \(\|x\|<1/2\) such that
\begin{equation*}
\|y-Tx\|_Y<\eps.
\end{equation*}
In particular, there is a \(z_1\in B_{0_X,\frac{1}{2}}\) such that \(\|y-Tz_1\|_Y<c/2\text{.}\) Similarly, there is a \(z_2\in B_{0_X,\frac{1}{4}}\) such that \(\|(y-Tz_1)-Tz_2\|_Y<c/4\text{.}\) Proceeding by induction, we build a sequence \(z_n\) with \(z_n\in B_{0_X,\frac{1}{2^n}}\) and \(\|y-T(z_1+\dots+z_n)\|_Y<c/2^n\text{.}\) It follows that the sequence \(x_n=z_1+\dots+z_n\) is Cauchy and so converges to some \(x\in X\text{.}\) Moreover,
\begin{equation*}
\|x\|=\|\sum_{n=1}^\infty x_n\|\leq\sum_{n=1}^\infty\|x_n\|<\sum_{n=1}^\infty2^{-n}=1
\end{equation*}
and, by construction,
\begin{equation*}
Lx=\lim Lx_n=y.
\end{equation*}
Hence,
\begin{equation*}
B_{0_Y,c}\subset L(B_{0_X,1})
\end{equation*}
Example 4.4.6. An open map.
\begin{equation*}
\frac{d}{dx}:C^1([0,1])\to C^0([0,1]).
\end{equation*}
It is continuous and surjective, so it is an open map.
\begin{equation*}
Tx=y.
\end{equation*}
Assume that the equation is solvable, namely that \(y\) is in the range of \(L\text{.}\) Then, the fact that \(L\) is an open map grants the solvability of \(Tx=y'\) for every \(y'\) close enough to \(y\text{.}\)
Corollary 4.4.7. The Bounded Inverse Theorem.
Let \(X,Y\) be two Banach spaces and let \(L\in\cB(X,Y)\) be a bijection. Then \(T^{-1}\in\cB(X,Y)\) as well.Lemma 4.4.8.
Let \(X\) be a Banach space with respect to both \(\|\cdot\|_1\) and \(\|\cdot\|_2\text{.}\) Then, if there is a \(c>0\) such that \(\|x\|_1\leq c\|x\|_2\text{,}\) the two norms are equivalent.Proof.
Theorem 4.4.9. The Closed Graph Theorem.
Let \(X,Y\) be two Banach spaces and let \(L\in\cL(X,Y)\text{.}\) Then \(L\in\cB(X,Y)\) if and only if its graph
\begin{equation*}
\Gamma_L = \{(x,Lx) : x\in X\}
\end{equation*}
is closed.Proof.
\begin{equation*}
\|x_n-x\|_2\to0.
\end{equation*}
By the Lemma above, \(\|\cdot\|_1\) and \(\|\cdot\|_2\) are equivalent, so that
\begin{equation*}
\|Lx\|_Y = \|x\|_2 - \|x\|_1 \leq \|x\|_2 \leq c\|x\|_1 = c\|x\|_X,
\end{equation*}
namely \(L\) is bounded. (\(\Leftarrow\)) Assume that \(L\) is bounded and let \((x_n,Lx_n)\to(x,y)\) a converging sequence in the graph. Then, by the continnuity of \(L\text{,}\) \(Lx_n\to Lx\text{.}\) Hence, \(y=Lx\) and so the graph is closed.Subsection 4.4.1 Fréchet spaces versions
Theorem 4.4.10. The Uniform Boundedness Principle.
Let \(X\) be a Fréchet space and \(Y\) a locally convex space (i.e. a non-complete version of a Fréchet space) and let \(\cF\subset \cB(X,Y)\) be a family of bounded operators. Then, if \(\cF\) is pointwise bounded, it is equicontinuous. Namely if, for every \(x\in X\text{,}\) the set \(\{Tx\,:\,T\in\cF\}\) is bounded in \(Y\text{,}\) then for every \(k\geq1\) there is a \(k'\geq1\) and a \(C>0\) so that
\begin{equation*}
q_k(Tx)<C p_{k'}(x)\text{ for all } x\in X\text{ and }T\in\cF.
\end{equation*}
Proof.
\begin{equation*}
U_m = \left\{x\in X\,:\,\sup_{T\in\cF} q_k(Tx)<m\right\}.
\end{equation*}
By hypothesis, each of those sup is finite and therefore
\begin{equation*}
\Cup_{i=1}^\infty U_m = X.
\end{equation*}
By Baire's theorem, there must be a \(U_{m_0}\) that contains some ball \(B_{x_0,r}\text{.}\) Hence, for some \(\eps>0\text{,}\) \(U_{m_0}\) contains some "seminorm tube" \(x_0+\{x\in X\,:\,p_n(x)<\eps\}\text{,}\) namely
\begin{equation*}
x_0 + x \in U_m\;\implies\;\sup_{T\in\cF} q_k(T(x_0+x))<m.
\end{equation*}
Using linearity and the triangular inequality we find ultimately that there is some \(M>0\) such that
\begin{equation*}
p_n(x) < \eps \; \implies \; \sup_{T\in\cF} q_k(T(x))<M.
\end{equation*}
Using the same arguments in Proposition 4.2.14, we get the equicontinuity.Theorem 4.4.11. The Open Mapping Theorem.
Let \(X,Y\) be Fréchet spaces and let \(L\in\cB(X,Y)\) be onto, i.e. \(L(X)=Y\text{.}\) Then \(L\) is an open map, i.e. \(L\) takes open sets to open sets.Proof.
Example 4.4.12. An open map.
\begin{equation*}
\frac{d}{dx}:C^\infty([0,1])\to C^\infty([0,1]).
\end{equation*}
It is not surjective because the derivative of all constant functions is zero. Now, set \(A=\{f\in C^\infty([0,1])\,:\,f(0)=0\}\) and denote by \(A\) the restriction of \(d/dx\) to \(A\text{.}\) Then \(A\) is continuous and surjective and so it is an open map.Corollary 4.4.13. The Bounded Inverse Theorem.
Let \(X,Y\) be two Fréchet spaces and let \(L\in\cB(X,Y)\) be a bijection. Then \(T^{-1}\in\cB(X,Y)\) as well.Theorem 4.4.14. The Closed Graph Theorem.
Let \(X,Y\) be two Fréchet spaces and let \(L\in\cL(X,Y)\text{.}\) Then \(L\in\cB(X,Y)\) if and only if its graph
\begin{equation*}
\Gamma_L = \{(x,Lx) : x\in X\}
\end{equation*}
is closed.Proof.
Example 4.4.15.
\begin{equation*}
\frac{d}{dx}:C^\infty([0,1])\to C^\infty([0,1]).
\end{equation*}
It is continuous and surjective, so it is an open map.