Fréchet Derivative Computation

Question

I’m trying to compute a Fréchet derivative, and I’m really getting bogged down in some details. I’d appreciate some help!

Let $U,V,E$ be Hilbert spaces, and let $f:U \rightarrow E$ and $g:V \rightarrow E$ be smooth (not necessarily linear) maps. I am trying to take the derivative $DF$ of the map $F: U \oplus V \rightarrow U$ given by: $$F(x,y) = (D_xf)^*(g(y) - f(x))$$ Where $(D_xf)^*: E \rightarrow U$ is the linear adjoint of $D_xf : U \rightarrow E$.

Here’s my attempt. I’m getting bogged down in some details with the second derivative. I’m pretty bad at analysis tbh.

Consider the quantity $Q = F(x + h_1, y + h_2) - F(x,y)$. We wish to find a bilinear map $A: U \times V \rightarrow U$ such that $A(h_1, h_2)$ is a good approximation of $Q$. We can expand

\begin{align*} Q & = (D_{x + h_1}f)^*(g(y + h_2) - f(x + h_1)) - (D_xf)^*(g(y) - f(x)) \\ & = [(D_{x + h_1}f)^*(g(y + h_2)) - (D_xf)^*(g(y))] - [(D_{x + h_1}f)^*(f(x + h_1)) - (D_xf)^*(f(x))] \\ & = Q_1 + Q_2 \end{align*}

We can now tackle these two pieces separately. But Here’s where I start running into trouble For $Q_1$, we compute \begin{align*} Q_1 & = (D_{x + h_1}f)^*(g(y + h_2)) - (D_xf)^*(g(y)) \\ & = (D_{x + h_1}f)^*(g(y + h_2)) - (D_{x}f)^*(g(y+ h_2)) + (D_{x}f)^*(g(y + h_2)) - (D_xf)^*(g(y)) \\ & = (D_{x + h_1}f - D_{x}f)^*(g(y + h_2)) + (D_xf)^*(g(y + h_2) - g(y)) \\ & \approx (D_{x + h_1}f - D_{x}f)^*(g(y + h_2)) + (D_xf)^* D_yg(h_2) \end{align*} I’m not sure how to handle the $(D_{x + h_1}f - D_{x}f)^*(g(y + h_2))$ term. Clearly it must have something to do with the second derivative $D^2_xf$, but I’m not sure how to express it. Similar difficulties happen with $Q_2$.

A push in the right direction would be appreciated! Thanks

Answer 1

Your map $F$ is of the form $T(x,y)[\phi(x,y)]=\text{ev}(T(x,y),\phi(x,y))$ for a smooth map $T:U\times V\to \text{Hom}(E,U)$ and $\phi:U\times V\to E$. Now, this is the perfect situation to apply the product rule (what you’re doing is trying to reprove the simple product rule and getting lost by the complexities of having specific but tedious functions). See here for the general product rule (which is really just a special case of the chain rule…).

Unwinding all of that in this situation gives \begin{align} DF_{(x,y)}[(h,k)]&=\left(D^2f_x(h,\cdot)\right)^*[g(y)-f(x)]+(Df_x)^*[Dg_y(k)-Df_x(h)], \end{align} i.e $DF_{(x,y)}$ is the linear map $U\times V\to U$ whose value on $(h,k)\in U\times V$ is given by the expression on the right.

Here’s the meaning of the adjoint on the first term: $f$ is a mapping $U\to E$, so the second derivative at a point $x$ is an element $D^2f_x\in \text{Hom}(U,\text{Hom}(U,E))\cong \text{Hom}^2(U;E)$. So, evaluating on $h\in U$ gives an element $D^2f_x(h,\cdot)\in \text{Hom}(U,E)$ and it is this linear map which is getting adjointed (and then being evaluated on $g(y)-f(x)$). And finally the reason why the derivative ‘goes through’ the adjoint is because adjointing is a linear map between certain $\text{Hom}$ spaces, and since the derivative at any point of a linear map is itself.

And if you don’t feel like explicitly writing the inputs $(h,k)$, you could write it as \begin{align} DF_{(x,y)}&=(D^2f_x\cdot dx)^*[g(y)-f(x)]+(Df_x)^*[Dg_y\cdot dy- Df_x\cdot dx], \end{align} where the $dx,dy$ can be treated as vector-valued $1$-forms and the $\cdot$ means evaluation… or simply you can think of them as placeholders for where the $h,k$ ought to go when making an evaluation.