Fréchet derivatives of $\sum_{n=1}^\infty x_n^2/n^3 -\sum_{n=1}^\infty x_n^4$

Question

I read that the second order Fréchet derivative $F''(0)$ of linear functional $F:\ell_2\to\ell_2$, where $\ell_2$ is the separable real Hilbert space, defined by $$F(x_1,\ldots):=\sum_{n=1}^\infty\frac{x_n^2}{n^3}-\sum_{n=1}^\infty x_n^4$$ is equal to series $2\sum_{n=1}^\infty \frac{h_n^2}{n^3}$. I would say that, to calculate $F''(0)$, which is such that forall $\varepsilon>0$ there exists a $\delta>0$ such that $$\|\Delta x\|_{\ell_2}<\delta\Rightarrow\|F'(\Delta x)-F'(0)-F''(0)\Delta x\|_{\ell_2^\ast}\le\varepsilon\|\Delta x\|_{\ell_2}$$Although the text that I am following does not give many examples of how to calculate Fréchet derivatives of maps defined on spaces other than $\mathbb{R}^n$, I have been able to see that $F'(0)=0\in\ell_2^\ast$, but, in order to verify what $F''(0)$ is, I would say that it is necessary to calculate $F'(\Delta x)$ first, but I have no idea about how one should handle such problems... How can $F'(\Delta)$ and $F''(0)$ be calculated? I $\infty$-ly thank you for any answer!

Answer 1

When dealing with functionals, I prefer to write the derivative as $F'(x;h)$, meaning $$F(x+h)=F(x)+F'(x;h)+o(\|h\|)\tag{1}$$ So, $F'(x;\cdot )$ is a linear functional on $\ell^2$. To find it formally (without justification), just differentiate $F$ coordinatewise: $$F(x;h) = \sum_{n=1}^\infty\frac{2x_n }{n^3} h_n-\sum_{n=1}^\infty 4x_n^3 h_n \tag{2}$$ Then you can prove that $(1)$ indeed holds by splitting the sum over $n$ into $n\le N$ and $n>N$; the latter is small when $N$ is large, the former is finite, so derivative works as in calculus.

For the second derivative, you need to differentiate $F(x;h)$ with respect to $x$, that is to obtain $$F'(x+k;h)=F'(x;h)+F''(x;h,k)+o(\|k\| \|h\|) \tag{3}$$ Again, the formal computation is not difficult: $$F''(x;h,k) = \sum_{n=1}^\infty\frac{2 }{n^3} h_n k_n- \sum_{n=1}^\infty 12x_n^2 h_n k_n \tag{4}$$ and the verification of (3) might be tedious but without surprises. Note that $F''(x)$ is a bilinear functional of $h$ and $k$. It is symmetric, which reflects the symmetry of mixed derivatives.