Computing the adjoint operator

Question

Let $H_1$ and $H_2$ be Hilbert spaces and let $u \in H_1$, $v \in H_2$. Define operator $A: H_1 \to H_2$ by $$Af := \left<u,f\right>v.$$ I want to compute the adjoint $A^*$ of $A$. I know the adjoint exists because $A$ is bounded. Here is my attempt. $$\left<Af,g\right> =\left<\left<u,f\right>v,g\right> = \left<u,f\right>\left<v,g\right> = \left<\left<v,g\right>u,f\right>.$$ The problem is that $f$ is on the second component of the inner product. Is there a way to move $f$ to the first entry and obtain an inner product of the form $\left<f,A^* g\right>$? I know taking conjugate is a common way to swap the components, but I don't know how to get rid of the conjugate on the right hand side.

Answer 1

If the scalar product is defined as anti-linear in the second argument, then $A$ is not a linear operator. This is because for $\lambda \in \mathbb{C}$ we would have in general $A(\lambda f) = \langle u, \lambda f\rangle v = \bar{\lambda} \langle u, f \rangle v \neq \lambda A( f)$.

So it must be the case here that the scalar product is anti-linear in the first argument.

In that case we have $$ \langle Af, g\rangle = \overline{ \langle u,f\rangle }\langle v,g \rangle = \langle f, u \rangle \langle v,g \rangle = \langle f, \langle v,g \rangle u \rangle $$ and so $A^* g = \langle v, g \rangle u $.

Your idea to compute the adjoint was correct, only the definition of the scalar product (or the exercise) seems to be mixed up here.