# Normal operator is invertible if and only if it is bounded below

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?

• 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. – Daniel Fischer Apr 15 '17 at 21:39

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.
• 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 $H$. – D_S Apr 15 '17 at 22:54