Prove differentiability of functional.

Question

In $C[0;1]$ space let's consider following functional: $$\phi(f) = \int_{0}^{1}(1+f(t))^{3}dt.$$ Prove differentiability of $\phi$ and find $\mathrm{D}\phi(f)$ for: $f(t)=0$, $f(t)=t$, $f(t)=\cos t$.
My attempt was to start with Gateaux derivative to estimate candidate for $\mathrm{D}\phi(f)$.
It's easy to see that $\phi(f+\varepsilon g) - \phi(f)= \int_{0}^{1}(1+f+\varepsilon g)^{3}-\int_{0}^{1}(1+f)^{3} = \varepsilon \int_{0}^{1}3(1+f)^2g + O(\varepsilon^2)$ This attempt guided us differentiability of $\phi$ and moreover exact form of derative i.e. $\mathrm{D}\phi(f)=\int_{0}^{1}3(1+f)^2$ (am I correct here?). So now we have (for example let's take third case): $\mathrm{D}\phi(\cos t) = \int_{0}^{1}3(1+\cos t)^2 dt$ and I have just to calculate value of integral or I've omitted some step? My questions may seem a little naive, but it's my first problem of that kind and I want to be sure about proof.

Answer 1

For every $u,h \in X:=C([0,1])$ we have \begin{eqnarray} \phi(u+h)&=&\int_0^1(1+u(t)+h(t))^3\,dt\\ &=&\int_0^1\left[(1+u(t))^3+3(1+u(t))^2h(t)+3(1+u(t))h^2(t)+h^3(t)\right]\,dt\\ &=&\int_0^1(1+u(t))^3\,dt+3\int_0^1(1+u(t))^2h(t)\,dt+\int_0^1\left[3(1+u(t))h^2(t)+h^3(t)\right]\,dt,\\ &=&\phi(u)+L_u(h)+\psi(u,h), \end{eqnarray} where $$ L_u(h)=3\int_0^1(1+u(t))^2h(t)\,dt,\quad \psi(u,h)=\int_0^1\left[3(1+u(t))h^2(t)+h^3(t)\right]\,dt. $$ It is clear that for each $u \in X$ the map $L_u: X\to\mathbb{R},\ h \mapsto L_u(h)$ is linea, and for every $u, h \in X$ we have $$ |L_u(h)| \le 3\int_0^1(1+u(t))^2|h(t)|\,dt \le \|1+u\|_\infty^2\|h\|_\infty, $$ i.e. $L_u$ is continuous. Since for every $u,h \in X$ we have $$ |\psi(u,h)|\le 3\int_0^1|1+u(t)||h(t)|^2+\int_0^1|h(t)|^3\,dt \le 3\|1+u\|_\infty\|h\|_\infty^2+\|h\|_\infty^3, $$ it follows that $$ \lim_{h\to 0}\frac{|\psi(u,h)|}{\|h\|_\infty}=0. $$ Hence $\phi$ is differentiable and $$ D\phi(u)[h]=L_u(h)=3\int_0^1(1+u(t))^2h(t)\,dt \quad \forall u,h \in X. $$