Showing that $\Psi(f) = \int^{b}_{a}\phi(f(x))dx$ is differentiable.

Question

Let $E = \mathcal{C}^0([a,b],\mathbb{R})$, provided with the $||\cdot ||_{\infty}$ norm. Let $\phi: \mathbb{R} \rightarrow \mathbb{R}$ that is $\mathcal{C}^1$. Show that the function given by $\Psi:E \rightarrow \mathbb{R}$: $$ \Psi(f) = \int^{b}_{a}\phi(f(x))dx $$ is differentiable.

I wasn't able to prove the statement, and I am not even sure if what I did is correct at all.

We have: $f$ is a continuous function from $[a,b]$ to $\mathbb{R}$, as $[a,b]$ is compact, then $f([a,b])$ is also a compact. As $\phi$ is $\mathcal{C}^1$, then as $f([a,b])$ is a compact, $\phi$ is Riemann integrable on $f([a,b])$. Now this shows that the function $\Psi$ is well defined.

In order to study the differentiability, I doubt that showing for $f,g \in E$, and $\lambda \in \mathbb{R}^*_+$, the limit as $\lambda \rightarrow 0$, $\frac{1}{\lambda}(\Psi(f + \lambda g) - \Psi(f))$ exists(necessary but not sufficient condition). But this problem makes me thing of the fundamental theorem of calculus, but unfortunately, it doesn't seem to me that I could apply it here. So I believe I gotta work with the composition of functions, but I am rather lost.

Answer 1

We have that \begin{align*} \Psi(f+h)&=\int_a^b\phi \circ (f+h) \\ &=\int_a^b\big(\phi(f(x))+\phi'(f(x))h(x)+\epsilon_{f(x)}(h(x))\big)dx \\ &=\Psi(f)+\int_a^b\phi'(f(x)) h(x)dx+\int_a^b\epsilon_{f(x)}(h(x))dx. \end{align*}

So the obvious candidate for the derivative at $f$ is $h \mapsto \int_a^b \phi' \circ f \cdot h$. Use that $f$ is continuous and $\phi$ is $C^1$ (and also that $[a,b]$ is compact, and $f([a,b])$ too) to arrive that $\epsilon_{f(x)} (h)$ is "uniform on $x$", and hence you will be able to bound that error to show that it is $o(h)$.

Answer 2

I guess that is to look for a linear operator $A_{f}:C^{0}[a,b]\rightarrow{\bf{R}}$ such that \begin{align*} \lim_{h\rightarrow 0}\dfrac{|\Psi(f+h)-\Psi(f)-A_{f}(h)|}{\|h\|_{\infty}}=0. \end{align*} Set $A_{f}(h)=\displaystyle\int_{a}^{b}\phi'(f(x))h(x)dx$, then \begin{align*} |\Psi(f+h)-\Psi(f)-A_{f}(h)|&=\left|\int_{a}^{b}[\phi((f+h)(x))-\phi(f(x))-\phi'(f(x))h(x)]dx\right|\\ &\leq\int_{a}^{b}|\phi'(\xi_{h}(x))-\phi'(f(x))|\cdot|h(x)|dx\\ &\leq\|h\|_{\infty}\int_{a}^{b}|\phi'(\xi_{h}(x))-\phi'(f(x))|dx, \end{align*} where $\xi_{h}(x)$ is chosen by Mean Value Theorem, so \begin{align*} \dfrac{|\Psi(f+h)-\Psi(f)-A_{f}(h)|}{\|h\|_{\infty}}\leq\int_{a}^{b}|\phi'(\xi_{h}(x))-\phi'(x)|dx\rightarrow 0 \end{align*} by Lebesgue Dominated Convergence Theorem if $h\rightarrow 0$ in $\|\cdot\|_{\infty}$.

Answer 3

Since $\phi$ is $C^1$ we can write $\phi(t+h) -\phi(t)-\phi'(t)h = \int_0^1 (\phi'(t+sh)-\phi'(t)) h ds$. Since $\phi'$ is uniformly continuous, for any $\epsilon >0$ we can find some $\delta>0$ such that if $|h| < \delta$, then $|\phi(t+h) -\phi(t)-\phi'(t)h| \le \epsilon |h|$ for all $x$.

In particular, if $\|\nu\|_\infty < \delta$, then $| \phi(f(x)+\nu(x)) -\phi(f(x))-\phi'(f(x))\nu(x) | \le \epsilon \|\nu\|_\infty $ for all $x$.

Integrating the expression $\phi(f(x)+\nu(x)) -\phi(f(x))-\phi'(f(x))\nu(x)$ gives $\|\nu\|_\infty < \delta$, then $|\Psi(f+\nu)-\Psi(f) -\int_a^b \phi'(f(x)) \nu(x) dx| \le (b-a) \epsilon \|\nu\|_\infty$.

In particular, this shows that the Fréchet derivative of $\Psi$ is $D \Psi(x)(\nu) = \int_a^b \phi'(f(x)) \nu(x) dx$.