If $T:X\to Y$ is a surjective linear map, then there is $M>0$ such that $\forall y,\exists x$ s.t. $y=T x,\|x\| \leqslant M \|y\|$

Question

Assume $X,Y$ are Banach spaces, $T \in L(X, Y)$, $\operatorname{Im}(T)=Y$.

How to prove \begin{array}{l} \exists M>0 \quad \text { s.t. } \forall y \in Y \text { , } \exists x \in X \\ \quad y=T x,\|x\| \leqslant M \cdot\|y\| \end{array}

I can only see that $T$ is surjective, so there must be an $x$ which satisfies '$Tx=y$'. But how to prove the norm issue, and the existence of the constant $M$?

Any suggestions shall be appreciated!

Answer 1

Since $T$ is a surjective continuous linear map between Banach spaces, it is an open map. Hence there is some $r>0$ such that $B_Y(0,r)\subseteq T(B_X(0,1)).$ Let $y\neq0\in Y$. Then $\frac{y}{2r\|y\|}\in B_Y(0,r).$ Hence there exists $z\in B_X(0,1)$ such that $T(z)=\frac{y}{2r\|y\|}.$ Let $x:=2r\|y\|z$. Then $T(x)=y$ and $$\|x\|=\|(2r\|y\|z)\|\leq 2r\|y\|\quad(\because \|z\|\leq 1).$$ Thus the claim holds with $M=2r$.