Derivative of adjoint operator-valued function

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?

• You can get proper double norm bars by using \| instead of ||. Jan 18 at 2:16
• This is a very subtle question!!!! Yes, seeing that the "strong" (as opposed to "uniform") operator topology is probably relevant for unbounded operators is (in my opinion) a very useful point. Jan 18 at 4:22
• @joriki Thanks, noted! Jan 18 at 7:16
• Dear @paulgarrett , thanks your your comment. Yes, but also in the case of a bounded operator-valued function this can make a difference; what I mean mean is that one could also define the derivative in terms of the SOT for bounded operator-valued functions; then their derivatives (if existent in the SOT) must not be bounded operators anymore. And so the derivative in the SOT might exist in cases where it does not exist for the uniform topology, which is important in the context of quantum mechanics, which also is the context where this question came up. Jan 18 at 9:19
• @Jakob, Aha! I was not aware of the significance of derivatives that exist in SOT but not in uniform! In my own business, repn theory and automorphic forms..., often integrals of operators converge in the SOT, but not the uniform... Jan 18 at 20:25

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}

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

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