# Derivative of $f \mapsto \int_{0}^l g(f(x))dx$

I'm looking to find the derivative (with respect to $$f$$) of

$$\int_{0}^l g(f(x))dx.$$

Assume of course that everything needed further is sufficiently smooth. Let us denote $$f \mapsto \int_{0}^l g(f(x))dx.$$In general, denoting $$G(f) = \int_0^l f(x)dx$$ would mean that $$G$$ is a linear functional, meaning the derivative of $$G$$ is just $$G$$ again. However, denoting $$H(f) = g \circ f$$ what I would then need is $$(G \circ H)'(f),$$ but I'm not sure how to proceed here, or even if I'm approaching this in the right manner. Would I need to turn to functional derivatives or am I (hopefully!) missing something easier here?

• Your mapping is from a function space to the real line. How do you even define the derivative? Dec 6, 2021 at 14:22

## 1 Answer

With derivative we mean the Fréchet-derivative. This is the most natural approach in this case, since in the $$\mathbb{R}^n$$ case, it coincides with the notion of derivative that is known from Calc 1/2.

Well, you have not specified the space in which $$f$$ lives. So let us assume that it is $$X:= L^2((0, l))$$. We need $$g \in C^1(\mathbb{R})$$ bounded with bounded derivative. Let $$h_k$$ be a non-zero sequence that converges to the zero function in $$X$$. We can w.l.o.g. assume that convergence is even pointwise a.e. (this is because convergent sequences in $$L^2$$ have a.e.-convergent subsequences). Then, let $$f \in X$$: $$G(f+h_k)-G(f) = \int^l_0 g(f(x)+h_k(x)) - g(f(x))~\mathrm{d}x = \int^l_0 \int^1_0 g'(f(x)+rh_k(x))h_k(x)~\mathrm{d}r~\mathrm{d}x$$ We can define the derivative in $$f$$ as a bounded functional $$DG_f: X \rightarrow \mathbb{R}$$ $$DG_f(h) := \int^l_0 g'(f(x))h(x)~\mathrm{d}x.$$ Note, that using the Hölder-inequality yields: $$\frac{1}{\lVert h_k \rVert_X}\lvert G(f+h_k) - G(h_k) - DG_f(h_k) \rvert = \frac{1}{\lVert h_k \rVert_X} \left \lvert \int^l_0 \int^1_0 h_k(x) (g'(f(x) + rh_k(x)) -g'(f(x))~\mathrm{d}r~\mathrm{d}x\right \rvert \leq \lVert h_k \rVert_X \frac{1}{\lVert h_k \rVert_X} \int^l_0 \int^1_0 \lvert g'(f(x) + rh_k(x))- g'(f(x)) \rvert^2~\mathrm{d}r~\mathrm{d}x$$ The integral converges to $$0$$ as $$k \rightarrow \infty$$ because of pointwise convergence of $$h$$ and the Lebesgue theorem. This proves Fréchet-differentiability.