T is a bounded linear operator and surjective, prove that there exists $k>0$ such that $\|x\|\leq k\|y\|$.

Question

$B(X,Y)$ denotes the space of bounded linear operators from $X$ to $Y$.

Question: Let $T\in B(X,Y)$ such that $T(X)=Y$. Show that there exists $k>0$ such that given $y\in Y$, there is an $x\in X$ such that $T(x)=y$ and $\|x\|\leq k\|y\|$.

Clearly, it is surjective. So we just need to prove the second part, i.e. prove that $\|x\|\leq k\|y\|$. I was thinking that I can use the corollary of Open Mapping Theorem to prove it:

Corollary: Let $X,Y$ be Banach spaces and let $T\in B(X,Y)$ be bijective. Then $T^{-1}\in B(X,Y)$ and there exist $c_1,c_2>0$ such that

$\|x\|\leq c_1\|Tx\|\leq c_2\|x\|,\;x\in X.$

Applying this corollary, we get $\|x\|\leq c_1\|y\|$.

But how can we prove $T$ is injective? Assuming that $Tx_1=Tx_2,\;\forall x_1,x_2\in X.$ How can we get $x_1=x_2$?

Could you please give me some ideas? Thank you so much.

Answer 1

Hint: from the Open Mapping theorem, there exists some $c>0$ such that every element of $Y$ with norm at most $c$ has a pre-image with norm at most $1$.

Answer 2

Hint: Consider the quotient space $\overline{X}=X/ker(T)$ , with the quotient norm, and the mapping $\overline{T}:\overline{X} \to Y$ defined by

$\overline{T}(\overline{x}):=Tx$, where $\overline{x}=x+ker(T).$

Then $\overline{T}$ is bounded and bijective.

Answer 3

You can't prove that it is injective. The only thing that you know about $T$ is that it's surjective, and there are linear surjective bounded operators between Banach spaces which are not injective. Take$$\begin{array}{ccc}\ell^1(\mathbb{N})&\longrightarrow&\mathbb R\\\displaystyle\sum_{n=1}^\infty a_n&\mapsto&a_1,\end{array}$$for instance.