Linear Transformation is Isomorphism iff Inverse Equals Adjoint

Theorem
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.

Let $U : \HH \to \KK$ be a bounded linear transformation.

Then the following are equivalent:


 * $(1): \quad U$ is an isomorphism
 * $(2): \quad U$ is invertible and $U^{-1} = U^*$, where $U^*$ denotes the adjoint of $U$.

$(1)$ implies $(2)$
Suppose that:


 * $U$ is an isomorphism.

That is:


 * $\innerprod {U g} {U h}_\KK = \innerprod g h_\HH$

for each $g, h \in \HH$.

Let $g, h \in \HH$.

From the definition of the adjoint, we have:


 * $\innerprod g h_\HH = \innerprod g {U^* U h}_\KK$

So, from Inner Product is Sesquilinear, we have:


 * $\innerprod g {h - U^* U h}_\HH = 0$

Setting $g = h - U^* U h$, we have:


 * $\innerprod {h - U^* U h} {h - U^* U h}_\HH = 0$

So from the positiveness of the inner product, we have:


 * $h - U^* U h = 0$

so:


 * $h = U^* U h$

for all $h \in \HH$.

So:


 * $U^* U = I_\HH$

where $I_\HH$ is the identity map on $\HH$.

From Hilbert Space Isomorphism is Bijection, we have that:


 * $U$ is a bijection.

So:


 * $U$ is invertible.

Let $U^{-1} : \KK \to \HH$ be the inverse of $U$.

Then, we have:

So:


 * $U$ is invertible with $U^{-1} = U^*$.

$(2)$ implies $(1)$
Suppose that:


 * $U$ is invertible with $U^{-1} = U^*$.

By the definition of an isomorphism, we aim to show that:


 * $\innerprod {U g} {U h}_\KK = \innerprod g h_\HH$

for each $g, h \in \HH$.

Let $g, h \in \HH$.

Then, we have:

So:


 * $U$ is an isomorphism.