open mapping theorem

[[concept]]

Open Mapping Theorem

Let $B_1,B_2$ be two Banach spaces and let $T\in \mathcal{B}(B_1,B_2)$ be surjective. Then $T$ is an open map.

ie for all open subsets $U\subset B_1$, we have $T(U)\subset B_2$ is open.

Proof

$B(0,1)=\{b\in B_1:||b||<1\}$, then $T(B(0,1))$ contains an open ball in $B_2$ centered at $0$. Then, since $T$ is linear, we can scale everything properly to show that for any $b_1\in U\subset B_1$ open, there exists $ϵ>0$ such that $B(T(b_1),ϵ)\subset T(U)$

We will first prove that if

Showing that the closure of the image $T(B(0,1))$ contains an open ball $B(0,r)$ $T$ surjective means everything in $B_2$ gets mapped to by something in $B_1$ ie $B_2={\bigcup }_{n\in \mathbb{N}}\overline{T(B(0,n))}$

So by the Baire Category Theorem, there exists some $n_0\in \mathbb{N}$ such that $\overline{T(B(0,n_0))}$ contains an open ball.

Now, since $T$ is a linear operator, $n_0\overline{T(B(0,1))}$ contains an open ball. ie, $\overline{T(B(0,1))}$ contains an open ball. Equivalently, there exists some $v_0\in B_2$ and $r>0$ such that $B(v_0,4r)\subset \overline{T(B(0,1))}$.

ie, we can pick some $u_1\in B(0,1)$ such that $v_1=T(u_1)\in T(B(0,1))$ such that $||v_0-v_1||<2r$

$4$ here to make some calculations easier later

we choose

Then $B(v_1,2r)\subset B(v_0,4r)\subset \overline{T(B(0,1))}$ If $||v||<r$, then $\frac{1}{2}(2v+v_1)\in \frac{1}{2}B(v_1,2r)\subset \frac{1}{2}\overline{T(B(0,1))}=\overline{T\left(B\left(0,\frac{1}{2}\right)\right)}$ And this means

v =-T\left( \frac{u_{1}}{2} \right) + \frac{1}{2}(2v + v_{1}) &\in -T\left( \frac{u_{1}}{2} \right) +\overline{T\left( B\left( 0, \frac{1}{2} \right) \right)} \\ &= \overline{T\underbrace{\left( -\frac{u_{1}}{2} + B\left( 0, \frac{1}{2} \right) \right)}_{\subset B(0,1) \text{ since } u_{1} \in B(0, 1)}}\\ &\subset \overline{T(B(0,1))} \\ \end{align}

ie, (in $B_2$), we have $B(0,r)\subset \overline{T(B(0,1))}$. This implies that $B(0,2^{-n}r)=2^{-n}B(0,r)\subset 2^{-n}\overline{T(B(0,1))}=\overline{T(B(0,2^{-n}))}$ for all $n\in \mathbb{N}$

Now we prove that $B\left(0,\frac{r}{2}\right)\subset T(B(0,1))$.

Showing that the image $T(B(0,1))$ contains an open ball $B\left(0,\frac{r}{2}\right)$

Let $||v||<\frac{r}{2}$. Then $v\in \overline{T\left(B\left(0,\frac{1}{2}\right)\right)}$ . This implies there exists some $b_1\in B\left(0,\frac{1}{2}\right)$ such that $||v-T{b}_1||<\frac{r}{4}$

ie, $v-T{b}_1\in \overline{T\left(B\left(0,\frac{1}{2}\right)\right)}⟹$ there is some $b_2\in B\left(0,\frac{1}{4}\right)$ such that $||v-T{b}_1-T{b}_2||=||v-T(b_1-b_2)||<\frac{r}{4}\cdot \frac{1}{2}=\frac{r}{8}$ continuing in this manner, we obtain a sequence $\{b_k{\}}_{k\in \mathbb{N}}\subset B_1$ such that

1. $||b_k||<2^{-k}$
2. $\left|\left|v-{\sum }_{k=1}^{n}T{b}_k\right|\right|<2^{-n-1}r$

The series ${\sum }_{k=1}^{\infty }{b}_k$ is absolutely summable (since the norm of the sums are bounded). And since $B_1$ is Banach, this means there exists some $b\in B_1$ such that $b={\sum }_{k=1}^{\infty }{b}_k$

Further, by the triangle inequality, we have $||b||={\lim }_{n\to \infty }\left|\left|{\sum }_{k=1}^n{b}_k\right|\right|\le {\lim }_{n\to \infty }{\sum }_{k=1}^{\infty }||b_k||$ And we can bound this $={\sum }_{k=1}^{\infty }||b_k||<{\sum }_{k=1}^{\infty }{2}^{-k}=1$ And since $T$ is a bounded linear operator (and thus continuous), we have $Tb={\lim }_{n\to \infty }T\left({\sum }_{k=1}^n{b}_k\right)={\lim }_{n\to \infty }{\sum }_{k=1}^{\infty }T{b}_k=v$ ie, $v\in T(B(0,1))$. Thus $B\left(0,\frac{r}{2}\right)\subset T(B(0,1))$ in $B_2$.

Thus, if $0$ was an interior point of $U$ in the domain, then it is still an interior point in the image of $U$.

Generalizing to any open set $U\subset B_1$ is open and $b_2=T{b}_1\in T(U)$. Then there exists some $ϵ>0$ such that $b_1+B(0,ϵ)=B(b_1,ϵ)\subset U$. And, from above, there exists some $\delta >0$ such that $B(0,\delta )\subset T(B(0,1))$. ie

Suppose

B(b_{2}, \epsilon\delta) &= b_{2}+\epsilon B(0, \delta) \\ &\subset b_{2} + \epsilon T(B(0, 1)) \\ &= T(b_{1}) + \epsilon T(B(0, 1)) \\ &= T(b_{1} + \epsilon B(0, 1)) \\ &= T(b_{1} + B(0, \epsilon)) \\ &\subset T(U) \end{align}$$ Since $b_{1}+B(0, \epsilon)$ is contained in $U$. Thus we've found a ball around any $b_{2} \in T(U)$.

\tag*{$\blacksquare$}

NOTE

closed graph theorem $⟺$ open mapping theorem

References

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