# Why is it that $\overline{\operatorname{Im}(\lambda - T)} = \operatorname{ker}(\overline{\lambda} - T^*)^\perp$?

Let $H$ a Hilbert space and $T: H \to H$ a self-adjoint, continuous operator. I try to understand why it is that $\overline{\operatorname{Im}(\lambda - T)} = \operatorname{ker}(\overline{\lambda} - T^*)^\perp$.

We will show that $\operatorname{Im}(\lambda - T)^\perp = \operatorname{ker}(\overline{\lambda} - T^*)$.

"$\supseteq$": Let $z \in \operatorname{ker}(\overline{\lambda} - T^*)$ and $y \in \operatorname{Im}(\lambda - T)$. Then there exists a $x \in H$ with $(\lambda - T)x = y$ and we can deduce

$$\langle z, y \rangle = \langle z, (\lambda - T)x \rangle = \langle (\overline{\lambda} - T^*) z, x \rangle = \langle 0, x \rangle = 0.$$ Hence we get $z \in \operatorname{Im}(\lambda - T)^\perp$.

"$\subseteq$": Let $z \in \operatorname{Im}(\lambda - T)^\perp$. Then we have for all $x \in H$ that $0 = \langle z, (\lambda - T) x \rangle = \langle (\overline{\lambda} - T^*) z, x \rangle$. Thus $(\overline{\lambda} - T^*) z = 0$ and it follows that $z \in \operatorname{ker}(\overline{\lambda} - T^*)$.

Finally we can deduce that $\overline{\operatorname{Im}(\lambda - T)} = \operatorname{ker}(\overline{\lambda} - T^*)^\perp$.

It is fairly easy to show (I omit the $\lambda$'s) $$\ker T^* = ({\rm im} \ T)^\perp$$ You may then (show and use) $(V^\perp)^\perp = \overline{V}$
Your proof has by the way a problem since the sequence $(x_n)$ need not be bounded (but you take the limit outside before turning $T$ into $T^*$).
• The two inclusions are fine. But you should add at least the statement that for any subspace $V$ one has: $\overline{V} = (V^\perp)^\perp$ (even better a proof or some reference to a proof if you have shown it in your course?) – H. H. Rugh Sep 14 '16 at 10:52