# Fréchet Derivative of functionals which depend on the derivative

Let $$\Omega:=[x_a,x_b]\times[t_a,t_b]\subset \mathbb{R}^2$$ and consider the functional $$D:C^2\left(\Omega,\mathbb{R}\right)\rightarrow \mathbb{R}$$ such that $$D[y]= \iint_{\Omega}d(x,t,y) \, dx \, dt$$ where $$d(x,t,y)= (1+y(x,t))^p\int_{x_a}^{x} \dfrac{dy(z,t)}{dt}\,dz$$ where $$p\in\mathbb{N}$$.

Is such a functional Fréchet-differentiable at any point $$y_0 \in C^2\left(\Omega,\mathbb{R}\right)$$? And how do we calculate its derivative?

In particular, I am not sure how the derivative in the integrand should be dealt with. When applying the limit definition, should I consider something like $$D[y_0+\eta]= \iint_\Omega\left[(1+y_0(x,t)+\eta(x,t))^p\int_{x_a}^{x} \left(\dfrac{dy_0(z,t)}{dt}+\dfrac{d\eta(z,t)}{dt}\right)dz\right]\, dx\,dt \quad?$$

• If $y$ were of class $\mathscr{C}^2,$ the standard product rules and differentiation under integral sign would apply. For a case like this, I believe you need to do the limit definition. – Will M. Jan 21 at 16:40
• @WillM. I tried to edit my question, making my point a bit clearer. I actually thought that $y$ being in $C^1$ or $C^2$ didn't make any difference, am I wrong? – William Tomblin Jan 21 at 16:55
• Isnt the integeral just y(b)-y(a)? – lalala Jan 21 at 17:00
• Isn’t $(1+y)^p$ a function, and not a scalar? – Mindlack Jan 21 at 17:49
• Then yes, your expression for $D[y_0+\eta]$ is correct. – Mindlack Jan 21 at 22:42

Let us consider the space $$\mathrm{X} = \mathscr{B}^1_{\mathbf{R}}(\mathrm{U} \times \mathrm{I})$$ of differentiable real valued bounded function with continuity defined on the product of the open set $$\mathrm{U}$$ of $$\mathbf{R}^d$$ and the compact interval $$\mathrm{I} = [a, b]$$ of $$\mathbf{R}.$$ Endow $$\mathrm{X}$$ with the structure of complete normed space by setting $$\|f\|_\mathrm{X} = \|f\| + \left\| \mathbf{D} f \right\|$$ (the sup-norm and the operator norm). (The fact that $$\mathrm{X}$$ is complete follows from the mean value theorem.)
Define $$\varphi:\mathrm{X} \to \mathrm{Y} = \mathscr{C}^b_\mathbf{R}(\mathrm{U}),$$ the right hand side is the space of continuous and bounded functions with the sup norm, by $$\varphi(f) = u_f, \quad u_f(x) = (1 + f(x))^p \int\limits_a^b \mathbf{D}_2f(x, t)\ dt.$$
Now, \begin{align*} u_{f + h}(x) &= (1 + f(x)+ h(x))^p \int\limits_a^b (\mathbf{D}_2f(x, t) + \mathbf{D}_2h(x, t))\ dt \\ &= \big[(1 + f(x))^p + p(1 + f(x))^{p-1} h(x) + o(1+f(x); h(x))\big]\\ &\times \int\limits_a^b (\mathbf{D}_2f(x, t) + \mathbf{D}_2h(x, t))\ dt \\ &= u_f(x)+p(1+f(x))^{p-1} h(x)\int\limits_a^b \mathbf{D}_2f(x, t)\ dt + (1 + f(x))^p \int\limits_a^b \mathbf{D}_2h(x, t)\ dt \\ &+ o(1+f(x); h(x))\int\limits_a^b (\mathbf{D}_2f(x, t) + \mathbf{D}_2h(x, t))\ dt, \\\\ \end{align*} where $$o(a; s)$$ is a sum of powers of $$s^\alpha,$$ where $$\alpha$$ runs from 2 until p, and the coefficients are powers of $$a$$ as well. Because $$f$$ is bounded, there exists a constant $$c = c(f)$$ such that $$\|o(1 + f; h)\| \leq c(\|h\|^2+\ldots+\|h\|^p) = o(\|h\|).$$ It is easy to check that the function $$L_f:h \mapsto L_f(h)$$ given according to the rule $$x\mapsto p(1+f(x))^{p-1} h(x)\int\limits_a^b \mathbf{D}_2f(x, t)\ dt + (1 + f(x))^p \int\limits_a^b \mathbf{D}_2h(x, t)\ dt$$ belongs to $$\mathrm{Y}$$ (that is, $$L_f:\mathrm{X} \to \mathrm{Y}$$) and it is linear. The desired derivative of $$\varphi$$ is therefore $$\mathbf{D}\varphi(f) = L_f.$$ Q.E.D.
• My problem contains yours, I believe. And I made a mistake too, instead of $(x)$ it should be $(x, t).$ – Will M. Jan 21 at 18:20