# Continuous operator between Banach spaces, closed range

I have some problems proving the following:

$T: X \rightarrow Y$ is a continuous, linear operator between Banach spaces.

Prove that $T$ is surjective $\iff$ $T^* : Y^* \rightarrow X^*$ is injective and $im T$ is closed.

This is my attempt:

Let's define $$S^{\perp} = \{ f \in X^* \ | \ f(s) = 0 \ \ \ \forall s \in S \}, \ S \subset X$$

$$^{\perp}V = \{ x \in X \ | \ f(x) = 0 \ \ \ \forall f \in V \}, \ V \subset X^*$$

Then evidently $S \subset ^{\perp}(S^{\perp})$ and $V \subset ( ^{\perp}V)^{\perp}$.

I don't see, however, how Hahn-Banach theorem implies that $S = ^{\perp}(S^{\perp}) \iff S$ is closed.$(*)$

If I prove that fact for S, I have that $\overline{imT}= ^{\perp}(kerT^*)$ and $\overline{imT^*} \subset (kerT)^{\perp}$

So the implication $\Leftarrow$ now follows from the first equation above.

Could you help me prove the other direction and tell me how to justify the fact $(*)$?

Thank you.

It is obvious that $S= {~}^\perp(S^\perp)$ implies that $S$ is a closed subspace. For the oposite implication, assume that $S$ is closed and $x\not\in S$. Let $T$ be the linear subspace spanned by $S$ and $x$. It is closed and each $y\in T$ is of the form $y=s+\alpha x$ for unique $s\in S$ and number $\alpha$. Define $\xi_0$ on $T$ by $\xi_0(s+\alpha x)=\alpha$. This is a bounded linear functional on $T$ and $S$ is in its kernel. Let $\xi$ be an extension of $\xi_0$ to the whole space. Then $\xi \in S^\perp$ and consequently $x\not\in {~}^\perp(S^\perp)$. This proves the nontrivial inclusion in $S= {~}^\perp(S^\perp)$.
Boundedness of $\xi_0$. Since $S$ is a closed subspace and $x\notin S$ the distance $d(x,S)=\inf\{ \| x-s\|; s\in S\}$ is a positive number, say $d(x,S)=\delta$. Let $y=s+\alpha x$ be an arbitrary vector in $T$. If $\alpha \ne 0$, then $$\| y\|=\| s+\alpha x\|=|\alpha|\| \frac{1}{\alpha}s+x\|\geq |\alpha|\delta=\delta|\xi_0(s+\alpha x)|=\delta|\xi_0(y)|.$$ It is obvious that $\delta|\xi_0(y)|=0\leq \|y\|$ if $y=s\in S$, i.e., if $\alpha=0$. Thus, in any case we have $|\xi_0(y)|\leq \frac{1}{\delta}\| y\|$ for any $y\in T$ which gives $\| \xi_0\| \leq \frac{1}{\delta}$.
• Thank you. Could you also help me with proving that if the operator $T$ is surjective, then $T^*$ is injective and $imT$ is closed? I can see that if $T$ is surjective, then $imT = Y$ and so $Y^{\perp} = ker(T^*)$. But I need to get that $ker(T^*) = (0)$ and $Y=^{\perp}ker(T^*)$ – Spencer Jan 13 '15 at 17:50
• If $\xi \in ker(T^*)$, then $T^*\xi=0$ and therefore $0=\langle x,T^*\xi\rangle=\langle Tx,\xi\rangle$ for any $x\in X$ which means that $im(T)\subseteq \ker \xi$. Since $T$ is surjective it follows that $\xi=0$. – Janko Bracic Jan 13 '15 at 17:56
• Oh, right. And the fact that the kernel is zero implies that $^{\perp}kerT^* = Y$. Thank you! – Spencer Jan 13 '15 at 18:10
• Could you tell me why $\xi _0$ is bounded? – Spencer Jan 17 '15 at 21:18