7

I'm confused about some equation I've seen in a book and want to write down some thoughts. I would appreciate, if somebody could tell me whether I'm terribly mistaken or not:

Let $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a Hilbert space over $\mathbb R$ and $f:H\to\mathbb R$ be Fréchet differentiable. By definition, the Fréchet derivative ${\rm D}f$ of $f$ is a mapping $H\to H'$$^1$. By Riesz' representation theorem, for each $L\in H'$ we can find a unique $v\in H$ with $$Lu=\langle u,v\rangle\;\;\;\text{for all }u\in H\;.$$ So, we should be able to identify ${\rm D}f$ with a mapping $H\to H$ such that for all $x\in H$ $${\rm D}f(x)u=\langle u,{\rm D}f(x)\rangle\;,\tag 1$$ where ${\rm D}f(x)$ is considered as an element of $H'$ and $H$ on the left-handed and right-handed side, respectively.

How would we state $(1)$ for the second derivative ${\rm D}^2f$?

$^1$ Let $H'$ denote the topological dual space of $H$.

0xbadf00d
  • 14,208
  • The second derivative at $x$ is a bilinear map $D^2f(x)\colon H\times H\to\mathbb{R}$, so you need some tensor product. – egreg Apr 23 '16 at 16:26
  • @egreg But we can consider ${\rm D}^2f(x)$ as an element of $\mathfrak L(H,\mathfrak L(H,\mathbb R))$ and write $${\rm D}^2f(x)(u,v)=\langle v,{\rm D}^2f(x)u\rangle;,$$ right? – 0xbadf00d Apr 23 '16 at 16:37
  • It's bad practise (and not fair with respect ot people who spend their time to answer your questions) to change a question after you received an answer in such a way that the answer becomes wrong or incomplete. Next time please consider creating a new question in case this is necessary. – Thomas Apr 23 '16 at 19:10
  • @Thomas I'm sorry, it wasn't my intent to "invalidate" your answer and I don't think that my edit did that. The paragraph I've added is of marginal importance for the question (and hence, I didn't want to create a new question for that). – 0xbadf00d Apr 23 '16 at 20:58

1 Answers1

4

This is what in finite dimension is called the gradient of $f$, if it exists (and which you may call gradient in this case as well). It's the same idea as in the finite dimensional case. In order for this to work you need a natural isomorphism between the vector space and it's dual (which you usually don't have but) which is induced by the scalar product in case of a Hilbert space.

Thomas
  • 23,023
  • How would we state $(1)$ (in the question) for the second derivative ${\rm D}^2f$? – 0xbadf00d Apr 23 '16 at 16:12
  • @0xbadf00d This is, in principle, the same idea, but we get into multilinear mappings here, which involves a bit algebra and in the infinite dimensional case additional complications, since a Hilbert space basis is not a basis in the sense of linear algebra. But roughly speaking, the second derivative is a bilinear map $H\times H rightarrow \mathbb{R}$, so if you freeze one argument for the map restricted to the other factor you can do the same thing. In finite dimensions this results in a matrix representation for the second derivative, in Hilbert space you'd expect a linear operator. – Thomas Apr 23 '16 at 17:38