In my notes I have the following theorem
If $T:X \to Y$ is a bounded operator where $X,Y$ are complex Hilbert spaces. Prove that $Im (T)$ is a closed set if and only if there exists positive constant $c$ such that $c \| x \| \leq \| Tx \|$ for $x \in (\ker T )^{\perp }$.
The proof of '$\impliedby$' I understand, but the proof my notes give for '$\implies$' is very brief and I don't understand.
Here's the proof.
Suppose we have the operator $T_1 : (\ker T)^{\perp } \to Im(T)$ such that $T_1x =Tx$, which is bijection and continuous. Then by the open mapping theorem, the desired inequality follows.
It's not obvious to me why the inequality follows. Also I can't see why $T_1$ is a bijection.
My understanding so far is this.
$T_1$ is continuous since $T$ is bounded.
$Im(T)$ is closed so $\overline{Im(T)} =Im(T)$, if it was true that $Im(T)$ is dense in $Y$ , we would get that $T_1$ is surjective, but I can't see if that's the case.
I don't understand why is $T_1$ injection.
Now for the inequality, since $T$ is continuous and bijection then $T^{-1}$ exists and is bounded by say $\frac{1}{c}$, so $$\| x \| =\| T^{-1} T x \| =\| T^{-1} \| \| Tx \| \leq \frac{1}{c} \| Tx \|.$$
So $c \| x \| \leq \| Tx \|$.
I also don't see where the open mapping theorem fits in.
Can you explain the proof?
Best Answer
Consider the map between Hilbert spaces ( the closedness of the image ensure completeness) $\operatorname{ker}(T)^{\perp} \to \operatorname{Im}(T)$. It is bijective and continuous, so it is bicontinuous (Banach open mapping theorem). You have to check that it is bijective ( use $X= \operatorname{Ker}(T) \oplus \operatorname{Ker}(T)^{\perp}$).