A Particular Frechet Derivative and Interpretation I would like find the Fréchet derivative of the following functional:
$$
\begin{align}
F : C[0,1] &\rightarrow \mathbb{R}\\
w &\mapsto \frac{\int_0^1 xw(x)f(x) \, dx}{\int_0^1 w(x)f(x) \, dx}.
\end{align}
$$
How can I do it?
Also, is that the correct way to think about how $\frac{E[xw(x)]}{E[w(x)]}$ would change if I were to perturb $w(x)$ a little? ($E$ is the expectation operator and $x$ is a continuously distributed random variable between 0 and 1 and $f$ is its density).
Thanks.
 A: This is an application of the chain rule for Frechet-Derivatives (see
e.g. http://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative#Properties).
Let us write
\begin{eqnarray*}
\Gamma_{1}: & C\left(\left[0,1\right]\right)\rightarrow\mathbb{R}, & w\mapsto\int_{0}^{1}w\left(x\right)\cdot xf\left(x\right)\, dx,\\
\Gamma_{2}: & C\left(\left[0,1\right]\right)\rightarrow\mathbb{R}, & w\mapsto\int_{0}^{1}w\left(x\right)f\left(x\right)\, dx.
\end{eqnarray*}
Assuming $f\in L^{1}\left(\left[0,1\right]\right)$ (this is true
in your application, since then $f\geq0$ with $\int_{0}^{1}f\left(x\right)\, dx=1$),
we easily see that $\Gamma_{1},\Gamma_{2}$ are bounded(!) linear(!)
functionals with $\left\Vert \Gamma_{1}\right\Vert ,\left\Vert \Gamma_{2}\right\Vert \leq\left\Vert f\right\Vert _{1}$.
In particular, $U:=\left\{ w\in C\left(\left[0,1\right]\right)\mid\Gamma_{2}\left(w\right)\neq0\right\} $
is open in $C\left(\left[0,1\right]\right)$.
This implies that also
$$
\Gamma:C\left(\left[0,1\right]\right)\rightarrow\mathbb{R}^{2},w\mapsto\left(\Gamma_{1}\left(w\right),\Gamma_{2}\left(w\right)\right)
$$
is a bounded linear map.
Their Frechet-derivative of this map is thus given by ``itself'',
i.e.
$$
\left(D\Gamma\right)\left(w\right)=\Gamma
$$
for all $w\in C\left(\left[0,1\right]\right)$. (If you don't know/believe
this, check it using the definition).
It is easy to see that the map
$$
\varrho:\mathbb{R}\times\mathbb{R}^{\ast}\rightarrow\mathbb{R},\left(x,y\right)\mapsto\frac{x}{y}
$$
is partially differentiable with continuous partial derivatives and
thus totally differentiable with
$$
\left(D\varrho\right)\left(x,y\right)=\left(\begin{array}[t]{cc}
\frac{1}{y} & -\frac{x}{y^{2}}\end{array}\right)
$$
which corresponds to the Frechet-derivative
$$
\left(\left(D\varrho\right)\left(x,y\right)\right)\left\langle \left(\begin{matrix}u\\
v
\end{matrix}\right)\right\rangle =\frac{u}{y}-\frac{x}{y^{2}}v,
$$
where I have written $A\left\langle x\right\rangle $ for the linear
map $A$ applied to $x$.
Now your function $F$ is given by
$$
F:U\rightarrow\mathbb{R},w\mapsto\varrho\left(\Gamma\left(w\right)\right).
$$
By the chain rule, $F$ is thus Frechet-differentiable with derivative
\begin{eqnarray*}
\left(\left(DF\right)\left(w\right)\right)\left\langle v\right\rangle  & = & \left(\left(D\varrho\right)\left(\Gamma\left(w\right)\right)\right)\left\langle \left(\left(D\Gamma\right)\left(w\right)\right)\left\langle v\right\rangle \right\rangle \\
 & = & \left(\left(D\varrho\right)\left(\Gamma\left(w\right)\right)\right)\left\langle \Gamma\left\langle v\right\rangle \right\rangle \\
 & = & \frac{\left(\Gamma\left\langle v\right\rangle \right)_{1}}{\left(\Gamma\left(w\right)\right)_{2}}-\frac{\left(\Gamma\left(w\right)\right)_{1}}{\left(\Gamma\left(w\right)\right)_{2}^{2}}\cdot\left(\Gamma\left\langle v\right\rangle \right)_{2}\\
 & = & \frac{\int_{0}^{1}v\left(x\right)\cdot xf\left(x\right)\, dx}{\int_{0}^{1}w\left(x\right)\cdot f\left(x\right)\, dx}-\frac{\int_{0}^{1}w\left(x\right)\cdot xf\left(x\right)\, dx}{\left(\int_{0}^{1}w\left(x\right)\cdot f\left(x\right)\, dx\right)^{2}}\cdot\int_{0}^{1}v\left(x\right)\cdot f\left(x\right)\, dx\\
 & = & \frac{\mathbb{E}\left(v\left(X\right)\cdot X\right)}{\mathbb{E}\left(w\left(X\right)\right)}-\frac{\mathbb{E}\left(w\left(X\right)\cdot X\right)}{\mathbb{E}\left(w\left(X\right)\right)}\cdot\mathbb{E}\left(v\left(X\right)\right),
\end{eqnarray*}
for $w \in U$ and $v \in C([0,1])$. Here, I have written $X$ for the random variable with density function $f$ in the last step.
One interpretation is now that
\begin{eqnarray*}
\frac{\mathbb{E}\left(\left(w+v\right)\left(X\right)\cdot X\right)}{\mathbb{E}\left(\left(w+v\right)\left(X\right)\right)} & = & F\left(w+v\right)\\
 & = & F\left(w\right)+\left(\left(DF\right)\left(w\right)\right)\left\langle v\right\rangle +o\left(v\right)\\
 & = & \frac{\mathbb{E}\left(w\left(X\right)\cdot X\right)}{\mathbb{E}\left(w\left(X\right)\right)}+\frac{\mathbb{E}\left(v\left(X\right)\cdot X\right)}{\mathbb{E}\left(w\left(X\right)\right)}-\frac{\mathbb{E}\left(w\left(X\right)\cdot X\right)}{\mathbb{E}\left(w\left(X\right)\right)}\cdot\mathbb{E}\left(v\left(X\right)\right)+o\left(v\right)
\end{eqnarray*}
holds for $w\in U$, $v\in C\left(\left[0,1\right]\right)$, where
$o\left(v\right)$ is a term with $\frac{\left|o\left(v\right)\right|}{\left\Vert v\right\Vert }\rightarrow0$
for $\left\Vert v\right\Vert \rightarrow0$.
EDIT: With regard to your second question (cf. the comments), my thoughts are the following:
The problem here seems to be (at first glance), that you have a quotient
of two integrals instead of just one integral. To get closest to the
form you search, you could write everything into one integral like
this:
\begin{eqnarray*}
 &  & \frac{F\left(w+\varepsilon v\right)-F\left(w\right)}{\varepsilon}\xrightarrow[\varepsilon\rightarrow0]{}\left(DF\right)\left(w\right)\left\langle v\right\rangle \\
 & = & \int_{0}^{1}\left(\frac{xf\left(x\right)}{\int_{0}^{1}w\left(x\right)\cdot f\left(x\right)\, dx}-f\left(x\right)\cdot\frac{\int_{0}^{1}w\left(x\right)\cdot xf\left(x\right)\, dx}{\left(\int_{0}^{1}w\left(x\right)\cdot f\left(x\right)\, dx\right)^{2}}\right)v\left(x\right)\, dx\\
 & = & \int_{0}^{1}\frac{f\left(x\right)}{\mathbb{E}\left(w\left(X\right)\right)}\left(x-\frac{\mathbb{E}\left(w\left(X\right)\cdot X\right)}{\mathbb{E}\left(w\left(X\right)\right)}\right)\cdot v\left(x\right)\, dx,
\end{eqnarray*}
which would suggest
$$
\frac{\delta F}{\left(\delta w\right)\left(x\right)}=\frac{f\left(x\right)}{\mathbb{E}\left(w\left(X\right)\right)}\left(x-\frac{\mathbb{E}\left(w\left(X\right)\cdot X\right)}{\mathbb{E}\left(w\left(X\right)\right)}\right).
$$
But I am note sure to what extend this is what you are looking for.
