Let $f:E\to \Bbb{R},\;p\mapsto f(p)=\int_{0}^{1}p^3{(t)}dt$. Then $f$ is differentiable and we can compute $f'(u)$

Question

I have been faced with this problem. Let $$f:E\to \Bbb{R}$$ $$p\mapsto f(p)=\int_{0}^{1}p^3{(t)}dt.$$ I want to prove that $f$ is differentiable and also compute $f'(u).$ $E=R_n[x]$ is provided with the following norm $\Vert p\Vert=\sup|p(t)|,\;\;\forall\;t\in [0,1]$

Here is what I've done:

$f(p) = \int_{0}^{1}p^3{(t)}dt, \tag 1$

we have

$f(p + h) = \int_{0}^{1}(p + h)^3{(t)}dt \tag 2$

for any $h \in E; \tag 3$

we may expand the right-hand side as follows:

$(p + h)^3 = (p + h)(p + h)(p + h) = (p + h)(p^2 + ph + hp + h^2)$ $= p^3 + p^2 h + php + ph^2 + hp^2 + hph + h^2 p + h^3; \tag 4$

thus

$$f(u + h) - f(u) = \int_{0}^{1}(p^2 h + php + ph^2 + hp^2 + hph + h^2 p + h^3){(t)}dt$$ $$= \int_{0}^{1}(p^2 h + php + hp^2 + ph^2 + hph + h^2 p + h^3){(t)}dt; \tag 5$$

$L(u)(h) = \int_{0}^{1}(p^2 h + php + hp^2){(t)}dt; \tag 6$

and

$$\Vert h\Vert \varepsilon(h)= \int_{0}^{1}( ph^2 + hph + h^2 p + h^3){(t)}dt $$

$$ \Vert \varepsilon(h)\Vert= \frac{\Vert \int_{0}^{1}( ph^2 + hph + h^2 p + h^3){(t)}dt \Vert}{\Vert h\Vert}. \tag 7$$

My question is: How do I force (7) to $0$? Please, can anyone help out?

Answer 1

Use $$\left|\int_a^b u\,\mathrm dt\right|\le (b-a)\|u\|$$ $$\|u+v\|\le\|u+v\|$$ and $$\|uv\|\le\|u\|\|v\|$$