In the following let $g(x,r,u)$ be a continuous function $g:[a,b]\times \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$. Furthermore we assume that the partial derivative $\frac{\partial g}{\partial u}$ exists and is also continuous. Now I want to calculate the Frechet derivative $DG$ of the operator $G: \mathcal{C}[a,b] \rightarrow \mathcal{C}[a,b]$ defined by:
$$G(u)(x) = u(x) – \int_{a}^{b} g(x,s,u(s)) ds$$
As a guess for $DG$ I use the Gateaux derivative of $G(u)(x)$. Which I calculate according to the definition as follows:
$$\frac{1}{t}\left(G(u+th)-G(u)\right) = h(x) – \frac{1}{t}\int_{a}^{b}g\left(x,s,u(s)+th(s)\right)-g\left(x,s,u(s)\right)$$
Now we can do a Taylor expansion of the integrand in the third argument:
$$g(x,s, z(t)) = g(x,s,z(0)) + \frac{\partial g(x,s,u(s))}{\partial z}\dot{z}(0) \left(z(t)-z(0)\right) + \mathcal{O}(t^{2})$$
Where $z(t) = u(s) + th(s)$. Using this expansion in the above integral we end up with the following expression for the Gateaux derivative:
$$\lim\limits_{t\to 0}\frac{1}{t}\left(G(u+th)-G(u)\right) = h(x) -\int_{a}^{b}\frac{\partial g}{\partial u}h(s)^{2} ds$$
Now we have to show that:
$$\lim\limits_{\lVert h \rVert \rightarrow 0}\frac{\lVert G(u+h) – G(u) – DG(u) \rVert}{\lVert h \rVert} = 0$$
Using the Gateaux derivative from above for $DG$ and working out the numerator I get:
$$\lVert G(u+h) – G(u) – DG(u) \rVert = \lVert \int_{a}^{b}\frac{\partial g}{\partial u} h(s)^{2} ds ~ – ~ \int_{a}^{b} \left(g\left(x,s,u(s)+h(s)\right) – g(x,s,u(s))\right) ds \rVert$$
For the first term I can use:
$\lVert \int_{a}^{b}\frac{\partial g}{\partial u} h(s)^{2} ds \rVert \leq \lVert h\rVert^{2} \int_{a}^{b}\frac{\partial g}{\partial u} ds$
And the fact the the partial derivative is continuous, hence bounded on $[a,b]$ am I right? But I can't see how to proceed with the second term. Above $\lVert . \rVert$ is the sup-norm. Is this the right way to calculate $DG$ or am I completely wrong with my approach?
Best Answer
The $h^2$ term is incorrect.
Note that $$\frac1t\int_a^b ds\ [g(x,s,u(s)+th(s)) -g(x,s,u(s))] =\int_a^b ds \frac1t\int_0^t dt\, \frac{\partial}{d\tau} g(x,s,u(s) +\tau h(s))$$ where $$\frac{\partial}{\partial \tau} g(x,s,u(s)+\tau h(s)) = g_u (x,s,u(s)+\tau h(s)) h(s) $$ Now note that by continuity of $g_u$ you have that $$\lim_{t\to0}\frac1t\int_0^tdt\ g_u(x,s,u(s)+\tau h(s)) = g_u(x,s,u(s))$$ and then the above expression converges to $$\int_a^b ds\ g_u(x,s,u(s))\,h(s)$$ and there is no square.
One final remark: This calculation shows that $\frac{G(u+th)-G(u)}{t}$ converges pointwise to $$x\mapsto h(x)- \int_a^b ds\ g_u(x,s,u(s))h(s).$$ But what you want is convergence in norm. I'm sure you can leverage the above calculation together with some continuity requirements on $g$ to get norm convergence.