Open Mapping Theorem

Definition

We say a linear operator $T:X\to Y$ is open if $\forall$ open set $U\subset X$ the set $T(U)=\{y\in Y:y=Tx\text{for some x}\in U\}$ is an open set in $Y$.

Proposition

$T:X\to Y$ is open iff $T(B(0,1))$ contains an open neighbourhood of 0.

Open Mapping Theorem

Let $X,Y$ be Banach spaces, $T:X\to Y$ be a bounded linear surjective operator. Then $T$ is an open map. (Drop either one of the completeness would fail.)

Proof
Claim:$\overline{T(B(0,1))}$ contains an open neighborhood of $0$.

$Y=T\left({\bigcup }_{n=1}^{\infty }B(0,n)\right)={\bigcup }_{n=1}^{\infty }T(B(0,n))$ Since $Y$ is complete, Baire category theorem says $\exists n\ge 1$ s.t. $\overline{T(B(0,n))}$ is not nowhere dense. $⟹\exists y_0\in Y,\exists r>0\text{ s.t. }\overline{T(B(0,n))}\supset B(y_0,r)$ As a result, $\overline{T(B(0,n))-T(B(0,n))}\supset B(y_0,r)-B(y_0,r)=\{y_1-y_2\text{ for }{y}_1,y_2\in \overline{B(y_0,r)}\}$ $\stackrel{\text{rescale}}{\to }\overline{T(B(0,2n))}\supset B(0,r)\stackrel{\text{rescale}}{\to }\overline{T(B(0,1))}\supset B(0,\frac{r}{2n})$ We now prove that $T(B(0,1))\supset B(0,\delta )$. Let $y\in B(0,\delta )\subset Y$. Want to solve $Tx=y$ with $\Vert x\Vert <1$. Let’s solve this approximately first. ($y\in T(B(0,1))$) From $B(0,\delta )\subset \overline{T(B(0,1))}$, $\exists x_1\in X,\Vert x_1\Vert <1$ s.t. $\Vert y-T{x}_1\Vert <\frac{\delta }{2}$. $⟹2(y-T{x}_1)\in B(0,\delta )$. Apply the same argument, not to $y$, but to $2(y-T{x}_1)$. So $\exists x_2\in X,\Vert x_2\Vert <\frac{1}{2}$ s.t. $\Vert 2(y-T{x}_1)-T{x}_2\Vert <\frac{\delta }{2}$. $⟹4(y-T{x}_1-\frac{1}{2}T{x}_2)\in B(0,\delta )$. Repeat $\dots \Vert y-T(x_1+\frac{1}{2}{x}_2)\Vert <\frac{\delta }{4}$. $\text{ we get }\{x_k{\}}_{k\in \mathbb{N}}\subset X,\Vert x_k\Vert <\frac{1}{2^{k-1}}\text{ s.t. }\Vert y-T\left(x_1+\frac{x_2}{2}+\cdots +\frac{x_k}{2^{k-1}}\right)\Vert <\frac{\delta }{2^k}\forall k\ge 1$. Take $n\to \infty$ : $y\to T({\sum }_{k=1}^{\infty }\frac{x_k}{2^{k-1}})$ and with $X$ complete, we get ${\sum }_{k=1}^{\infty }\frac{x_k}{2^{k-1}}$ converges. Let $x={\sum }_{k=1}^{\infty }\frac{x_k}{2^{k-1}}$ then $\Vert x\Vert <1$ and $Tx=y$.

Bounded Inverse theorem

Let $X,Y$ be Banach spaces, $T:X\to Y$ a bounded linear bijection, then the inverse map is also bounded i.e. $\exists c>0$ s.t. $\Vert T^{-1}y{\Vert }_X\le c\Vert y{\Vert }_Y\forall y\in Y$