Let $X,Y$ be Banach spaces and $T:X\to Y$ be a bounded linear operator. It is required to show that there is a constant $m>0$ such that $\|T(x)\|\geq m\|x\|$ for all $x\in X$ if and only if $T$ is injective and $T(X)$ is closed.

I proved the forward implication using the fact that $T^{-1}$ exists and it is bounded if and only if there is $m>0$ such that $m\|x\|\leq\|T(x)\|$ for all $x\in X$, which I have already proved.

However I cannot find a way to prove the other direction. I tried showing that $T(X)=Y$ which is just some wishful thinking so that I can invoke the above result. But I couldn't and I am not sure if it really is the case that $T(X)=Y$. So could someone please give me a hint? Thanks.

Edit: (-added later-) ($\Rightarrow$) $x_1\neq x_2\implies \|x_1-x_2\|>0\implies \|T(x_1)-T(x_2)\|=\|T(x_1-x_2)\|\geq m\|x_1-x_2\|>0\implies T(x_1)\neq T(x_2)$. Hence $T$ is injective.

Let $y\in \overline{T(X)}$. Then choose $(y_n)=(T(x_n))$ in $T(X)$ such that $y_n\to y$. For each $m,n\in\mathbb{Z^+},\|y_m-y_n\|=\|T(x_m)-T(x_n)\|=\|T(x_m-x_n)\|\geq m\|x_m-x_n\|$, and therefore since $(y_n)$ is cauchy $(x_n)$ is Cauchy and hence converges to some $x\in X$ as $X$ is complete. Now by continuity of $T$ we have $y=T(x)\in T(X)$. Hence $T(X)$ is closed.

($\Leftarrow$) $T:X\to T(X)$ is a bijective bounded linear operator and $T(X)$ is complete as it is closed. By open mapping theorem $T$ is an isomorphism and hence there exists $m>0$ such that $\|T(x)\|\geq m\|x\|$ for each $x\in X$.

@JohnMa Is the added later part alright?

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    $\begingroup$ What do you mean by $T^{-1}$ exists? In general $T$ is not surjective and $T^{-1}$ is not defined. $\endgroup$ – user99914 Sep 22 '17 at 19:52
  • $\begingroup$ @JohnMa I agree. I've made a mistake there. But could you give me a hint because I'm stuck. $\endgroup$ – Janitha357 Sep 22 '17 at 19:59

To give a hint:

$(\Rightarrow)$ Injectivity is easy. to show that $T(X)$ is closed, let $y\in \overline{T(X)}$. Then there is a sequence $y_k = T(x_k)$ so that $y_k \to y$. Try to use the inequality to show that $x_k \to x$.

$(\Leftarrow)$ This is indeed open mapping theorem (or it's corollary): The mapping $ T: X\to T(X)$ between Banach spaces is boundedly invertible.

  • $\begingroup$ could you please check the "added later" part in my post? $\endgroup$ – Janitha357 Sep 22 '17 at 20:45
  • $\begingroup$ @Janitha357: that looks perfect to me. $\endgroup$ – user99914 Sep 22 '17 at 20:47
  • $\begingroup$ Thank you once again. $\endgroup$ – Janitha357 Sep 22 '17 at 20:51

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