Surjective bounded linear map bound

Question

Let $X,Y$ be Banach spaces.

Let $T: X \rightarrow Y$ be a surjective bounded linear map.

Show that there is a constant $M>0$ such that for each $y\in Y $ there is a solution to $Tx=y$ with $\| x\| \leq M \| y\|$.

Let $B_Y(0,r)= \{ y \in Y : \| y \| < r\}$.

I have shown that there exists $r>0$ such that $B_Y(0,r) \subset T(B_X(0,1))$.

Not sure how to conclude from here.

Thank you in advance!

Note that the property you want to prove is not the same as "bounded below". — Martin Argerami, Feb 01 '21 at 05:03

score 2 · Accepted Answer · answered Feb 01 '21 at 01:22

2

The open mappin theorem implies that there exits $r>0$ such that $B_Y(0,r)\subset T(B_X(0,1))$. Let $y\in Y, y\neq 0$, ${r\over {\|y\|}}y\in B_Y(0,r)$ implies that there exists $x'\in B_X(0,1)$ such that $T(x')={r\over{\|y\|}}y$. We have $T({{\|y\|}\over r}x')=y$ and $\|{{\|y\|}\over r}x'\|\leq {{\|y\|}\over r}$. Take $M={1\over r}$.

answered Feb 01 '21 at 01:22

Tsemo Aristide

87,475

@Tselmo, I think your modified $y$ is not in the open ball. – Ruy Feb 01 '21 at 02:30

Surjective bounded linear map bound

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