2
$\begingroup$

Let $H$ be a Hilbert space, and let $T$ be a bounded linear operator on $H$. We say that $T$ is bounded below if there exists an $M > 0$ such that $\|Tx\| \geq M \|x\|$ for all $x \in H$.

It is a consequence of the open mapping theorem that $T$ is invertible if and only if it is bounded below and has dense image.

If $T$ is normal ($TT^{\ast} = T^{\ast}T$), I have heard that $T$ is invertible if and only if it is bounded below. Is this true?

$\endgroup$
  • $\begingroup$ For normal operators, we have $\lVert T^{\ast} x\rVert = \lVert Tx\rVert$ for all $x$, so if $T$ is normal and bounded below, $T^{\ast}$ is also bounded below, in particular injective. $\endgroup$ – Daniel Fischer Apr 15 '17 at 21:39
3
$\begingroup$

If $T$ is a normal operator, then we have

$$\lVert T^{\ast} x\rVert^2 = \langle T^{\ast} x, T^{\ast} x\rangle = \langle x, TT^{\ast} x\rangle = \langle x, T^{\ast} T x\rangle = \langle Tx, Tx\rangle = \lVert Tx\rVert^2$$

for all $x$, so if $T$ is normal and bounded below, we have

$$(\operatorname{im} T)^{\perp} = \ker T^{\ast} = \ker T = \{0\},$$

i.e. the image of $T$ is dense. Since $T$ is bounded below, the image is also closed. Hence $T$ is invertible.

$\endgroup$
  • $\begingroup$ That makes sense, thanks very much $\endgroup$ – D_S Apr 15 '17 at 22:52
  • $\begingroup$ As a note to myself when I read this later, if $W$ is a subspace of a Hilbert space $H$, then $W^{\perp} = \overline{W}^{\perp}$, and if $W$ is closed, then $H = W \oplus W^{\perp}$. It follows that if $W^{\perp} = 0$, then $W$ is dense in $H$. $\endgroup$ – D_S Apr 15 '17 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.