• $H$ 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 $n\in\left\{1,\ldots,d\right\}$.

  • $\begingroup$ $$ v=\lim_{t\to 0}\dfrac{f(x+te_n)-f(x)}{t}$$ $\endgroup$ – 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.

  • $\begingroup$ 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 $\sum_iv_ie_i=v$? $\endgroup$ – 0xbadf00d May 24 '16 at 18:17
  • $\begingroup$ 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. $\endgroup$ – s.harp May 24 '16 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.