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Let $X = \mathcal{C} \left( [0,1] \right)$ be the Banach space of continuous functions on $[0,1]$ (with the supremum norm) and define a map $F : X \rightarrow X$ by $$F(f)(x) = \int^{x}_{0} \cos(f(t)^{2})dt, x \in [0,1].$$

Show that $F$ is Fréchet differentiable and compute the Fréchet derivative $DF|_{f}$ for each $f \in X.$

So far I have the following.

Using the identity $\cos(A+\varepsilon B)=\cos(A)-\varepsilon B\sin(A)+\mathcal{O}(\varepsilon^2).$ We have \begin{align*}F(f+h)(x) &=\int^{x}_{0} \cos((f(t)+h(t))^{2})dt \\ &=\int^{x}_{0}\cos(f^2(t))-h(t)[2f(t)h(t)]\sin(f^2(t))+\mathcal{O}(h^2(t))dt \\ &=F(f)(x)+\int^{x}_{0}-2f(t)(h(t)\text{sin}(f^2(t))-h^2(t)\sin(f^2(t))+\mathcal{O}(h^2(t))dt \\ \end{align*} Let $T(h)=\int^{x}_{0}-2f(t)(h(t)\sin(f^2(t))dt$. I am able to show that $T(h)$ is a linear map. Then we have $$F(f+h)(x)-F(f)(x)=T(h)+\int^{x}_{0}\mathcal{O}(h^2(t))dt $$ I am confused about the Big-$\mathcal{O}$ notation and I do not understand what it would mean to integrate over it. Also, am I on the right track? Thank you in advance for any help provided.

Stackman
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1 Answers1

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I think your derivative $T$ is correct. In my experience big-$O$ notation is useful for obtaining the solution for this kind of problems, but for actual proofs it is often easier to write down explicit inequalities. So let's start doing exactly that.

$\renewcommand{\epsilon}{\varepsilon}$ By the error theorem for Taylor expansion $\cos(A + x) = \cos A - x \sin A - \frac12x^2 \cos \alpha$ for some $\alpha \in (A, A+x)$. Since bounding $\cos$ by $1$ we get $$ |\cos(A + x) - (\cos A - x \sin A)| \leq \frac12 x^2 $$

Now let $f, h \in \mathcal{C}[0, 1]$ such that $\|h\|_\infty < \epsilon$ for some $\epsilon > 0$. Then $$ \begin{align} |F(f + h)(x) - F(f)(x) - T(x) | \\ &\hspace{-45mm}= \left|\int_0^x \cos\left((f(t)^2 + h(t)^2)\right) - \cos\left(f(t)^2\right) +2 f(t) h(t) \sin\left(f(t)^2\right) \,dt \right|\\ &\hspace{-45mm}= \left|\int_0^x \cos\left((f(t)^2 + h(t)^2)\right) - \cos\left(f(t)^2\right) + \left( 2 f(t) h(t) + h(t)^2 \right) \sin\left(f(t)^2\right) \\ - h(t)^2 \sin\left(f(t)^2\right)\,dt\right| \\ &\hspace{-45mm}\leq \int_0^x \frac12\left| 2 f(t) h(t) + h(t)^2 \right|^2 + \left| h(t)^2 \sin\left(f(t)^2\right) \right| \,dt \\ &\hspace{-45mm}\leq \frac12 \left( 2 \|f\|_\infty \epsilon + \epsilon^2 \right)^2 + \epsilon^2 \end{align} $$

This last bound is obviously $O(\epsilon^2)$.

Carlos Esparza
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