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Let

  • $E$ be a normed space
  • $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a Hilbert space
  • $f:E\to H$ be Fréchet differentiable
  • $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ and $$f_n:=\langle f,e_n\rangle\;\;\;\text{for }n\in\mathbb N$$

How can we compute the second Fréchet derivative ${\rm D^2}f_n:E\to\mathfrak L(E,\mathfrak L(E,H))$ of $f_n$?$^1$

Let $$L_n:=\langle\;\cdot\;,e_n\rangle\;\;\;\text{for }n\in\mathbb N\;.$$ Then, since each $L_n$ is an element of $\mathfrak L(H,\mathbb R)$$^1$, $${\rm D}L_n(u)=L_n\;\;\;\text{for all }u\in H\text{ and }n\in\mathbb N$$ and hence $${\rm D}f_n(x)={\rm D}(L_n\circ f)(x)={\rm D}L_n(f(x))\circ{\rm D}f(x)=L_n\circ{\rm D}f(x)=\langle{\rm D}f(x),e_n\rangle$$ by the chain rule, for all $x\in E$ and $n\in\mathbb N$.

So, I guess that $${\rm D^2}f_n(x)={\rm D}({\rm D}f_n)(x)={\rm D}(L_n\circ{\rm D}f)(x)={\rm D}L_n({\rm D}f(x))\circ {\rm D^2}f(x)=L_n\circ{\rm D^2}f(x)=\langle{\rm D^2}f(x),e_n\rangle\;,$$ again by the chain rule, for all $x\in E$ and $n\in\mathbb N$.

But I'm unsure whether I made a mistake or not.


$^1$ Let $\mathfrak L(A,B)$ be the space of bounded, linear operators from $A$ to $B$.

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Yes, it is fine if you interpret $\langle D^2 f(x) , e_n \rangle$ as the bilinear form $$(y,z) \mapsto \langle D^2 f(x)[y,z], e_n \rangle.$$ As you already said, this follows simply from the chain rule and the linearity of $L_n$.

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