Ytukyg

I just started learning the Frechet Derivatives.
So I have a function $H:mathbb{R}^{Ntimes n}tomathbb{R}^{Ntimes n}$, i.e. $U^Tinmathbb{R}^{Ntimes n}$ and
$$H(U^T)=GWtimes (F(U))^T+Stimes U^T+C$$
with $G,W,Sin mathbb{R}^{N times N}$ are two matrices of size $Ntimes N$, $F(cdot)in mathbb{R}^ntomathbb{R}^n $ is a nonlinear function which maps each column vetor of $U$ to the corresponding column vector of $F(U)$, and $Cinmathbb{R}^{Ntimes n}$.

My question is what property should the nonlinear unknown function $F(cdot)$ satisfy to ensure the function $H(cdot)$ is Frechet differentiable? What does the Frechet derivative matrix looks like? What should I start?Thank you!

I'll rewrite the definition of $H$ as
$$
H(X) = GW F(X^T)^T + SX + C.
$$
Let's assume that $F$ is Frechet differentiable at a particular point $X^T$,
so that
$$
F(X^T + Delta X^T) = F(X^T) + F'(X^T) Delta X^T + e(Delta X),
$$
and the error term $e(Delta X)$ satisfies
$$
lim_{Delta X to 0} frac{|e(Delta X)|}{| Delta X |} = 0.
$$
Notice that
begin{align}
H(X + Delta X) &= GW F(X^T + Delta X^T)^T + SX + SDelta X + C \
&= GW left( F(X^T) + F'(X^T) Delta X^T + e(Delta X) right)^T + SX + S Delta X + C \
&= underbrace{GWF(X^T)^T + SX + C}_{H(X)} + underbrace{GW(F'(X^T) Delta X^T)^T + S Delta X}_{H'(X) Delta X} + underbrace{GW e(Delta X)^T}_{text{small}}.
end{align}

Comparing this with the equation
$$
H(X + Delta X) approx H(X) + H'(X) Delta X
$$
suggests that $H$ is differentiable at $X$ and that $H'(X)$ is the linear transformation defined by
$$
tag{1} H'(X) Delta X = GW(F'(X^T) Delta X^T)^T + S Delta X.
$$
To prove that this is true, we only need to show that
$$
tag{2} lim_{Delta X to 0} frac{| GW e(Delta X)^T |}{ | Delta X |} = 0
$$
To establish (2), let $L$ be the linear transformation defined by
$$
L(v) = GW v^T.
$$
Then
begin{align}
frac{| GW e(Delta X)^T |}{ | Delta X |}
&= frac{| L(e(Delta X)) |}{| Delta X |} \
&leq frac{| L | |e(Delta X) |}{| Delta X |}
end{align}
which approaches $0$ as $Delta X to 0$.

In order to reach the conclusion that $H$ is differentiable at $X$, we needed to assume that $F$ is differentiable at $X^T$.

I don't see a simpler way to express $H'(X)$, but maybe somebody else will.

They gave you the hand-wavy proof above. I am giving you the high-end proof now.

Recall from basic differential calculus:

Basic differentiation algebra: the derivative acts linearly $(f+ alpha g)'(x) = f'(x) + alpha g'(x)$; the derivative of constant functions is zero; and the derivative of continuous linear functions are themselves $f'(x) cdot h = f(h)$ whenever $f$ is linear and continuous.

Chain rule: if $g$ and $f$ are two functions defined on open subsets of normed vector spaces such that $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$ then the composite function $g circ f$ is differentiable at $c$ and its derivative is the composite of the derivatives $$(g circ f)'(x) = g'(f(x)) circ f'(x).$$
Abridged proof. Write $y = f(x)$ and $f(x + h) = f(x) + underbrace{f'(x) cdot h + o(h)}_k$ and $$g(y + k) = g(y) + g'(y) k + o(k) = g(y) + g'(y) cdot f'(x) cdot h + underbrace{g'(y) o(h) + o(k)}_{o(k)}. square$$

To your exercise. The function $H$ is differentiable at every $U$ where the function $F$ is differentiable as well.

Proof. The functions $varphi:V mapsto GW V^intercal$ and $psi = U mapsto SU^intercal$ are linear while the function $U mapsto C$ is contant. Therefore, the function $H = varphi circ F + psi + C$ will be differentiable at all points where $F$ is differentiable (by the chain rule) and its derivative is simply $$H'(U) = varphi'(U) circ F'(U)^intercal + psi'(U) = varphi circ F'(U) + psi.$$

If you are dealing with finite dimensional vector spaces, find bases of each so that (by denoting $[ cdot ]$ the matrix represantion) we get $$[H'(U)]=[varphi][F'(U)]^intercal + [psi]. square$$

Ammend. If the function $varphi$ is invertible then the differentiability of $H$ implies that of $F$ for we can write $F = varphi^{-1} circ (H - psi - C),$ and $varphi^{-1}$ being linear and continuous, it is differentiable. $square$