Let $T \in B(X,Y)=\{T \in \mathcal{L}(X,Y) | T$ is bounded $\}$ with $X$ and $Y$ Banach space such that it exist $\delta >0$ , $\|Tx\| \geq \delta \|x\|$ for all $x \in X$. Show that $T$ is an isomorphism of X in a closed subspace of $Y$
I need some help to solve this. I dont know where to begin with.
What I know about isomorphism
- maybe find T $\in \mathcal{L}(X,Y)$ Bijectif with $\|T\|=\|x\|$
- $T:X \rightarrow Y$ isomorphe iff Ker T=0 and ImT=Y
- $X \rightarrow X^*$ is isomorphic if X is an Hilberspace
What else I know is since T is bounded then T is continuous $\implies$ kerT is closed.
Suppose that $Tx = Ty$. Then $$ 0 = \Vert Tx-Ty \Vert \geq \delta \Vert x - y \Vert \quad \Rightarrow \quad \Vert x-y \Vert \leq 0 \quad \Rightarrow \quad \Vert x - y \Vert = 0, $$ which is possible if and only if $x = y$. Hence $T$ is injective.
Suppose now that $(Tx_n)_{n=1}^\infty \subset R_T$ is a convergent sequence in the range of $R_T$. Then it is in particular Cauchy, i.e. $$ \Vert Tx_n - Tx_m \Vert \to 0 , \quad n,m \to \infty. $$ Then $$ \Vert x_n-x_m \Vert \leq \frac{1}{\delta} \Vert Tx_n-Tx_m \Vert \to 0, \quad n,m\to\infty, $$ so that $(x_n)_{n=1}^\infty \subset X$ is Cauchy. Since $X$ is a Banach space, this sequence converges to some $x\in X$. Then by the boundedness of $T$, $$ Tx_n \to Tx \in R_T, \quad n \to \infty. $$ Hence, the range of $T$ is closed.
Thus $T:X\to R_T$ is an isomorphism between $X$ and the closed subspace $R_T$ of $Y$.