Banach fixed point theorem (application)

Question

Let $X$ and $Y$ be Banach Spaces, and $T: X \to Y$, where T is continuous, linear and bijective, and let $S: X \to Y$ (where $S$ is continuous and linear) with $|S|\cdot|T^{-1}|< 1$. Show that $S+T$ is bijective. (Hint: Use Banach fixed point theorem)

This question was in my test today. I live in Brazil, so sorry for my English.

Answer 1

We have that $$ S+T=T(I+T^{-1}S). $$ Define $$ K=\left(\sum_{n=0}^\infty (-1)^n (T^{-1}S)^n\right)T^{-1}. $$ Then, the series above converges as $\|T^{-1}S\|<1$, and hence $K$ is a bounded linear operator, and is readily shown that $$ K(S+T)=I.$$

Answer 2

If $X=Y$ in your question, we have \begin{equation*} S+T: X \rightarrow X. \end{equation*}

$S+T$ is a injective map, because $(S+T)(x) = 0 \Rightarrow x= 0 $.

Now, we want to show that for all $y \in X$, exists $x \in X$ such that $(S+T)(x)=y$. See that, \begin{equation*} (S+T)(x) = y \Leftrightarrow S(x) +T(x) =y \Leftrightarrow T^{-1}(y) - T^{-1}S(x) = x \end{equation*} Set, $A(x) = T^{-1}(y) - T^{-1}S(x)$. It's easy to show that $A$ is a contraction, then $S+T$ is surjective.