Some Main Theorems on NLS - 5. Banach Open Mapping Theorem

A map \(T:V \to W\) is said to be a open map if $T$ maps open sets in $V$ to open sets in $W$. It is a closed map if $T$ maps closed sets in $V$ to closed sets in $W$.

Theorem 2.5.1

Let $V$ and $W$ be Banach spaces. Let $T:V \to W$ be linear and continuous. Then $T$ is an open map if and only if $T$ is onto.

Proof. Let $U = B(0,\,1)$ be the open unit ball in $V$. We will show that there exists $\rho >0$ such that $B(0,\, \rho) \subseteq T(U)$. Then it follows that $\Omega$ is open in $V$ implies $T(\Omega)$ is open in $W$. Let $V = \bigcup_{k=1}^\infty kU$. Then,

\[ W = T(V) = T\left( \bigcup_{k=1}^\infty kU \right) = \bigcup_{k=1}^\infty T(kU). \]

By Baire Category Theorem, there exists $k \in N$ such that $\mybar{T(kU)}$ has nonempty interior. So, $B(y_0,\, \eta) \subseteq \mybar{T(kU)}$ for some $y_0 \in W$ and $\eta>0$. Then, ■

Example 2.5.2

Let $T : \R^2 \to \R$ by $T(x,\,y) = x$. Then, $T$ is linear, continuous, and onto. Thus, it is a open map.

However, $T$ is not a closed map. Take $C = \set{(x,\,y)}{y = \frac{1}{x}}$. Then $C$ is closed, whereas $T(C) = (0,\,\infty)$ is not.

Remark. If $T:V \to W$ is linear and open, then it is onto. \\

\textbf{Proof.} Take $w \in W$. Look at $U = B(0,\,1)$ in $V$. Since $T$ is open, $T(U)$ is also open in $W$ and contains $0$. Thus, $B(0,\,\epsilon) \subset T(U)$ for some $\epsilon>0$. Now, take $\lambda = \epsilon/2\norm{w}>0$. Then $\lambda w \in B(0,\,\epsilon) \subseteq T(U)$. Thus, there exists $u \in U$ such that $T(u) = \lambda w$. Hence, $T(\frac{u}{\lambda}) = w$.

Corollary 2.5.3

Let $V$ and $W$ be Banach. Suppose $T:V \to W$ is one-to-one, onto, linear, and continuous. Then $T^{-1} : W \to V$ is linear and continuous.

Proof. $T^{-1}$ is continuous if and only if $(T^{-1})^{-1}(U)$ is open in $V$ for all open set $U$ in $W$. However, since $(T^{-1})^{-1}(U) = T(U)$ is open by Banach Open Mapping Theorem, the proof is established. ■

Corollary 2.5.4

Let $\norm{\cdot}$ and $\opnorm{\cdot}$ be two complete norms on the same vector space $V$. Suppose there exists $\alpha>0$ such that

\[ \norm{x} \leq \alpha \opnorm{x} \qquad \text{for all $x \in V$}. \]

Then there exists $\beta>0$ such that

\[ \opnorm{x} \leq \beta \norm{x} \qquad \text{for all $x \in V$}. \]

Thus, the two norms are equivalent.

Proof. Let $T : (V,\, \opnorm{\cdot}) \to (V,\, \norm{\cdot})$ be the identity map, i.e., $T(x) = x$ for all $x \in V$. Then $T$ is bijection, linear, and $\norm{x} \leq \alpha \opnorm{x}$ for all $x \in V$. Thus, $T$ is continuous. Now, applying Corollary ??, we see that $T^{-1} : (V,\, \norm{\cdot}) \to (V,\, \opnorm{\cdot})$ is also continuous. Hence, there exists $\beta>0$ such that $\opnorm{x} \leq \beta \norm{x}$ for all $x \in V$. ■

Remark. Let $(V,\,\norm{\cdot})$ be a finite normed linear space and let $\{e_1,\, \ldots,\, e_n\}$ be a basis of $V$. Then for $x = \sum_{i=1}^{n} x_i e_i \in V$,

\[ \norm{x} \leq \sum_{i=1}^{n} \abs{x_i} \norm{e_i} \leq \left( \max_{1\leq i\leq n} \norm{e_i} \right) \sum_{i=1}^{n} \abs{x_i} = \alpha \norm{x}_1. \]

Since $(V,\, \norm{\cdot})$ and $(V,\, \norm{\cdot}_1)$ are Banach, these two norms are equivalent. Consequently, any two norms on $V$ are equivalent.

Example 2.5.5

Let $ \begin{aligned}[t] V &= l_1(\R) \\ & = \set{x = (x_1,\, x_2,\, \ldots)}{\norm{x}_1 = \textstyle \sum_{i=1}^{\infty} \abs{x_i} < \infty}. \end{aligned} $

Note that it is a Banach space. Now, put a new norm on $V$ by $\norm{x}_2 = \left( \sum_{i=1}^{\infty} \abs{x_i}^2 \right)^{1/2}$. Then we know that

\[ \left( \sum_{i=1}^{n} \abs{x_i}^2 \right)^{\frac{1}{2}} \leq \sum_{i=1}^{n} \abs{x_i} \leq \sum_{i=1}^{\infty} \abs{x_i} < \infty. \]

Letting $n \to \infty$, we have $\norm{x}_2 \leq \norm{x}_1$ for all $x \in V$.

Now, both $(V,\, \norm{\cdot}_1)$ and $(V,\, \norm{\cdot}_2)$ form normed linear spaces. Do we have $\norm{x}_1 \leq \beta \norm{x}_2$ for some $\beta>0$? Suppose the inequality holds. Take $x^{(k)} := (1,\, \tfrac{1}{2},\, \tfrac{1}{3},\, \ldots,\, \tfrac{1}{k},\, 0,\, 0,\, \ldots) \in V$. Then

\[ \norm{x^{(k)}}_1 = \sum_{n=1}^{k} \frac{1}{n} \leq \beta \left( \sum_{i=1}^{n} \frac{1}{n^2} \right)^{\frac{1}{2}} \leq \beta \left( \sum_{i=1}^{\infty} \frac{1}{n^2} \right)^{\frac{1}{2}} < \infty, \]

for all $k \in \N$. Taking limit as $k \to \infty$, we have $\infty < \infty$, a contradiction. It turns out that $(V,\, \norm{\cdot}_2)$ is not complete.

Definition 2.5.6

Let $(M,\,d)$ and $(N,\,\delta)$ be two metric spaces. For a function $f:M \to N$, the graph is

\[ G(f) := \set{(x,\,f(x))}{x \in M} \subseteq M \times N. \]

Define $\rho = d + \delta$. Note that $(M \times M,\, d + \delta)$ is a metric space where

\[ \rho((x_1,\, y_1),\, (x_2,\, y_2)) := \text{d}(x_1,\,x_2) + \delta(y_1,\,y_2). \]

If $G(f)$ is closed in this metric space, we say that $f$ has closed graph. Note that $G(f)$ is closed in $M \times N$ if $x_n \to x$ in $M$ and $f(x_n) \to y$ in $N$ implies $y = f(x)$, i.e., $(x,\,y) \in G(f)$.

Remark. Recall that the continuity of $f$ is defined as:

\[ x_n \to x \text{ in } M \ \Rightarrow \ f(x_n) \to f(x) \text{ in } N. \]clearly, `$f$ is continuous' implies `$G(f)$ is closed'.

Example 2.5.7

Consider $D: C_\R^1[a,\,b] \to C_\R[a,\,b]$ by $D(f) = f'$ where

\[ C_\R^1[a,\,b] := \set{f \in C_\R[a,\,b]}{f' \in C_\R[a,\,b]}. \]

Note that the two spaces both carry sup-norm.

Now, we claim that $D$ has closed graph, but NOT continuous.

Proof. To show $G(D)$ is closed, we need check that

\[ \left. \begin{aligned} f_n &\to f \text{ in } C_\R^1[a,\,b] \\ D(f_n) &\to g \text{ in } C_\R[a,\,b] \end{aligned} \ \right\} \ \Rightarrow \ g= D(f). \]

Assume $f_n \to f$ in $C_\R^1[a,\,b]$ and $D(f_n) \to g$ in $C_\R[a,\,b]$. Then

\[ \int_{a}^{t} f'_n(x) \text{d}{x} \to \int_{a}^{t} g(x) \text{d}{x} \quad \text{as $n \to \infty$}. \]

Since $f_n(t) - f_n(a) \to f(t) - f(a)$, we get

\[ f(t) - f(a) = \int_{a}^{t} g(x) \text{d}{x} \]

for $t \in [a,\,b]$. Differentiate both sides to get $f'(t) = g(t)$ for all $t \in [a,\,b]$.

Theorem 2.5.8 (Closed Graph Theorem)

Let $(V,\, \norm{\cdot})$ and $(W,\, \norm{\cdot})$ be Banach spaces. Let $T : V \to W$ be linear. If $T$ has a closed graph, then $T$ is continuous.

Proof. Define $\opnorm{x} := \norm{x} + \norm{T(x)}$ for $x \in V$. Clearly, this is a norm on $V$. We now show that $\opnorm{\cdot}$ is a complete norm on $V$. Let $\opnorm{x_n - x_m} \to 0$ as $n,\,n \to \infty$. Then,

\[ \norm{x_n - x_m} + \norm{T(x_n) - T(x_m)} \to 0. \]

Thus, both $\norm{x_n - x_m} \to 0$ and $\norm{T(x_n) - T(x_M)} \to 0$. As $V$ and $W$ are complete, $x_n \to x$ in $V$ and $T(x_n) \to y$ in $W$ for some $x \in V$ and $y \in W$. Since $T$ is a closed graph, we have $y = T(x)$. Thus,

\[ \opnorm{x_n - x} = \norm{x_n - x} + \norm{T(x_n) = y} \to 0. \]

Hence $x_n$ converges in $(V,\, \opnorm{\cdot})$ proving that $(V,\, \opnorm{\cdot})$ is a complete.

Now, $(V,\, \norm{\cdot})$ and $(V,\, \opnorm{\cdot})$ are Banach and $\norm{x} \leq \opnorm{x}$ for all $x \in V$. Applying a corollary of the Open Mapping Theorem, there exists $\alpha>0$ such that $\opnorm{x} \leq \norm{x}$ for all $x \in V$. Thus,

\[ \norm{T(x)} \leq \alpha \norm{x} \quad \text{for $x \in V$}. \]

Hence, $T$ is continuous. ■