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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?

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

1 Answer 1

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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.

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