Representation of the Fréchet derivative of $〈f,e_n〉$, where $f:H→H$, $H$ is a Hilbert space and $(e_n)_{n∈ℕ}$ is an orthonormal basis of $H$

Let

• $H$ be a $\mathbb R$-Hilbert space
• $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$
• $f:H\to H$ be Fréchet differentiable and $$f_n:=\langle f,e_n\rangle\;\;\;\text{for }n\in\mathbb N$$
• $\mathfrak L(A,B)$ denote the space of bounded, linear operators between normed $\mathbb R$-vector spaces $A$ and $B$ and $\mathfrak L(A):=\mathfrak L(A,A)$

Let $L_n:={\rm D}f_n(x)$ denote the Fréchet derivative of $f_n$ at $x\in H$ for some $n\in\mathbb N$. Then, $L_n$ is an element of $\mathfrak L(H,\mathbb R)$ $\Rightarrow$ $\exists!v\in H$ with $$L_n=\langle\;\cdot\;,v\rangle\tag 1$$ by Riesz' representation theorem. On the other hand, $$L_nu=\langle\underbrace{{\rm D}f(x)}_{=:\;L}u,e_n\rangle\;\;\;\text{for all }u\in H\;.\tag 2$$ Thus, by definition of the adjoint $L^\ast$, $$v=L^\ast e_n\;.\tag 3$$

Now, the concrete shape of $L^\ast$ is not obvious to me. In particular, $L^\ast$ is defined to be $v$, but $v$ is unknown. So, the question is: Is there some more concrete representation of $L^\ast$?

Note that we can find a concrete representation of $L^\ast$ when $H=\mathbb R^d$ for some $d\in\mathbb N$: In that case we obtain $$L(u)=\sum_{i=1}^du_i\frac{\partial f_n}{\partial x_i}(x)=u\cdot\nabla f_n(x)\;\;\;\text{for all }u\in H$$ and hence $$v=\nabla f_n(x)\;,$$ if $n\in\left\{1,\ldots,d\right\}$.

• $$v=\lim_{t\to 0}\dfrac{f(x+te_n)-f(x)}{t}$$ – Yiorgos S. Smyrlis May 20 '16 at 16:03

I'm not entirely sure what you are asking, maybe what I am writing does not give you what you are looking for. But there is essentially no difference from the finite dimensional case.

If you write $v=\sum_i v_i e_i$ then

$$L_n(e_i)=\langle e_i, v\rangle = v_i$$

And you have $v=\sum_i L_n(e_i)\ e_i$, which is the same as your equation $\nabla f_n(x) = v$. That this sum is well defined follows from $L_n$ being bounded linear functional, ie $v$ being in $H$, which is something of a circular argument.

If you denote $\nabla_j f(x):=Df(x)e_j$, then this becomes

$$v=\sum_j \langle \nabla_j f(x), e_n \rangle e_j$$

So $L^*(w)=\sum_j \langle \nabla_j f(x), w \rangle e_j$. I don't know if this is more pleasing.

• Do we really need to argument that the sum $\sum_i L_n(e_i)\ e_i$ is well-defined by noting that $L_n$ is a bounded, linear operator? Shouldn't that simply follow by your equation $L_n(e_i)=v_i$ and the well-definedness of $\sum_iv_ie_i=v$? – 0xbadf00d May 24 '16 at 18:17
• Yeah, thats what I meant with circular, the fact that $v$ exists follows from $L_n$ being bounded and the Riesz representation argument and the sum is well defined because its the components of $v$. On the other hand $L_n$ being bounded also gives you $L_n(u)=\sum_i u_i L_n(e_i)$ converges absolutely for any $u \in \mathscr l^2$, which makes $L_n(e_i)$ be square summable and thus $v=\sum_i L_n(e_i) e_i$ is a well defined vector. – s.harp May 24 '16 at 18:34