Derivative of adjoint operator-valued function

Question

Consider an infinite dimensional complex Hilbert space $H$. I think that for a bounded operator-valued function $A: x\mapsto A(x) \in \mathcal B(H)$, where $x\in \mathbb R$, we can define the derivative $A^\prime(x)$ as the (unique) operator which obeys

$$\lim\limits_{h\to 0} \left\|A^\prime(x) - \tfrac{A(x+h)-A(x)}{h}\right\|_{\mathrm{op}} =0 \quad, \tag 1$$

if the limit exists. Here $\|\cdot\|_{\mathrm{op}}$ denotes the operator norm.

Now I think that from $\|A(x)^*\|_{\mathrm{op}}=\|A(x)\|_{\mathrm{op}}$, where $^*$ denotes the adjoint, we can show that the derivative of $A^*: x\mapsto A(x)^*$ exists at $x$ and is simply the adjoint of the derivative of $A$ at $x$, i.e we have $$ (A^\prime(x))^* = (A^*)^\prime (x) \quad .\tag{2}$$

Question:

Can we find the same result as in $(2)$ if we define the derivative of bounded operator-valued functions in the strong operator topology instead of the uniform topology $(1)$? Or is there a weaker but similar result under some conditions?

Answer 1

Here is a partial answer for an admittedly very restrictive special case.

To start, consider the following subspace: $$\mathcal D(A^\prime(x)):=\left\{\psi\in H\,|\, \exists\, \psi_x \in H\, : \lim\limits_{h\to 0}\,\left\|\psi_x-\tfrac{A(x+h)-A(x)}{h}\psi \right\|=0\right\} \quad . $$

We find that for each $\psi \in \mathcal D(A^\prime(x))$, the corresponding $\psi_x$ is unique and this in turn allows us to define an operator $A^\prime(x): \mathcal D(A^\prime(x)) \longrightarrow H$ as $$A^\prime(x)\psi:=\psi_x$$ and extend by linearity. It is a well-defined linear operator and we call it the derivative of $A$ at $x$.$^1$

Now assume that $\mathcal D(A^\prime(x))=H$. I think that from the uniform boundedness principle it then follows that $A^\prime(x)$ is bounded$^2$, which in turn implies that $(A^\prime(x))^*$ is bounded and defined on the whole Hilbert space as well. With that we can show that $A^*: x\mapsto A(x)^*$ is differentiable in the weak sense and its derivative is $(A^\prime(x))^*$.

Indeed, by making use of the continuity of the underlying inner product $\langle \cdot,\cdot\rangle: H\times H \longrightarrow \mathbb C$, we find for all $\phi,\psi \in H$: \begin{align} \left\langle (A^\prime(x))^*\phi,\psi\right\rangle &= \langle \phi,A^\prime(x) \psi\rangle \\ &= \lim\limits_{h\to 0}\, \left\langle \phi,\tfrac{A(x+h)-A(x)}{h}\psi\right\rangle \\ &= \lim\limits_{h\to 0}\, \left\langle \left(\tfrac{A(x+h)-A(x)}{h}\right)^* \phi,\psi\right\rangle \\ &= \lim\limits_{h\to 0}\left\langle \tfrac{A^*(x+h)-A^*(x)}{h}\phi,\psi\right\rangle \quad . \end{align}

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$^1$ I think this resembles the notion given here, p. 20-21.

$^2$ Thanks to @V. Moretti for pointing that out.