Consider $H$ some euclidean Hilbert space, a $C^2$ function $f\colon H\to\mathbb{R}$, and $x,y\in H$. In the context of some problem about convexity and positive definiteness of the second derivative I wanted to write down, properly, as an exercise, the second derivative of the map $g:=f\circ m$, where $$m\colon [0,1]\to H,\; t\mapsto x +t(y-x),$$ as an element of $\mathcal{L}^2(\mathbb{R},\mathbb{R})$

So, for $t\in[0,1]$ it is $Dg(t)= Df(m(t))\circ m'(t)$ where $Df(m(t))\in\mathcal{L}(H,\mathbb{R})$ and $m'(t)\in\mathcal{L}(\mathbb{R},H)$ and using the gradients definition via Riesz it is $$Dg(t)=\langle m'(t)(\,\cdot\,),\nabla f(m(t)\rangle\in\mathcal{L}(\mathbb{R},\mathbb{R}),$$ Consider the function $\nabla f\colon H\to H$. Now, $$D^2g(t) = \langle m'(t)(\,\cdot\,), D(\nabla f\circ m)(t)\rangle + \langle m''(t)(\,\cdot\,),\nabla f(m(t))\rangle = \langle m'(t)(\,\cdot\,), D(\nabla f\circ m)(t)\rangle$$

It is $D(\nabla f\circ m)(t)= D(\nabla f)(m(t))\circ m'(t)$, so, what is $$D(\nabla f)(m(t))\in\mathcal{L}(H,H)?$$

Again via Riesz, there is a bilinear form $b\in\mathcal{L}^2(H,\mathbb{R})$ such that $$\langle D(\nabla f)(m(t))x,y\rangle =b(x,y),\; x,y\in H.$$

What is $b$? Is $b=D^2f(m(t))$? Is $D(\nabla f)(m(t))$ known as something like $\nabla^2 f(m(t))$? (Of course I know the Hessian matrix in the finite case.) Did I get the spaces right? Can I simplify further?

  • $\begingroup$ @CrazyIvan Sorry, please see the edit! And yes, I used the fact that the second derivative of $m$ in $t$ is zero for calculating $D^2g(t)$. $\endgroup$ – Ramen Dec 17 '17 at 10:50

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