TITLE: Let $T$ be a surjective linear map such that $ \|Tx\| \geq \frac{1}{2017} \|x\|$ for all $x \in X$. Show that $T$ is bounded.
QUESTION [5 upvotes]: Problem: Let $T$ be a surjective linear map from a Banach space $X$ onto a Banach space $Y$ such that $$ \|Tx\| \geq \frac{1}{2017} \|x\|$$ for all $x \in X$. Show that $T$ is bounded.
I thought of proving by contradiction: saying that if $T$ is not bounded then $T$ is not continuous (since we know that $T$ is bounded is equivalent to $T$ is continuous), but it get me to nowhere.
Could anyone give me a tip or a solution for this problem? Thanks

REPLY [8 votes]: After all the comments made I have this, hope it is right:
Let $T$ be a surjective map $T : X \to Y$. It s obvious that $T$ is injective since $\|Tx\|=0$ imply $\|x\|=0$ therefore $T^{-1}$ exists and 
$$\|T  T^{-1}x\|=\|x\|\geq \frac{1}{2017} \|T^{-1}x\|$$ or
$$\|T^{-1}x\|\leq 2017\|x\|$$ which means that $T^{-1}$ is bounded.
Now we are using The Open Mapping Theorem Corollary " Let $X$ and $Y$ be Banach spaces and $T \in \mathcal{L}(X,Y)$ be one-to-one and onto. Then $T^{-1}$ is continuous"
By this $T=(T^{-1})^{-1}$ is continuous and therefore $T$ is bounded.