I'm working on the following problem related to proving Frechet differentiability in $C^0$, but I'm not too comfortable with the whole $O(||h||)$ step of the process and was hoping someone could provide a work-through to clarify the logic.
Consider $J:C^0[0,2]\to R$ defined by $$J(x)=\int_0^2 (3t^5-12t^2x(t)+3tx(t)^2+2x(t)^3)dt$$ Show that $J$ is Frechet differentiable and give its Frechet derivative $Df(x)$
So, the first steps I find quite straightforward $$J(x+h)=\int_0^2 (3t^5-12t^2(x+h)+3t(x+h)^2)+2(x+h)^3))dt$$ So, due to $$(x+h)^2=x^2+2xh+h^2,\quad(x+h)^3=x^3+3x^2h+3xh^2+h^3\to$$ $$J(x+h)=\int_0^2(3t^5-12t^x-12t^h+3tx^2+6txh+3th^2+2x^3+6x^2h+6xh^2+2h^3)dt$$ Let's try to separate out $J(x)\to$ $$J(x+h)=\underbrace{\int_0^2(3t^5-12t^2x+3tx^2+2x^3)dt}_{=J(x)}+...$$ $$...+\int_0^2(-12t^2h+6txh+3th^2+6x^2h+6xh^2+2h^3)$$ Now, we need to know if this last line $=DJ(x)h+O(||h||)$. We quite easily see that $$J(x+h)=J(x)+\underbrace{\int_0^2(-12t^2h+6txh+6x^2h)dt}_{=DJ(x)h}+\int_0^2(3th^2+6xh^2+2h^3)dt$$ Thus, the focus just remains on the last remaining term. I believe the requirement is to prove that $$\lim_{||h||_{L_p}\to 0}\frac{\cdot}{||h||}_{L_p}=0$$ where $\cdot$ is just the left over terms from before. My first stumbling point is simply which p-norm do we have in $C^0$, is it the maximum-norm? If so, could anyone provide the final logic to prove differentiability?
Thank you very much
You indeed raise a very good point... with what norm are you endowing $\mathcal C^0([0,2])$?
Suppose that you’re using the sup norm $\Vert \cdot \Vert_\infty$. Then $$\begin{aligned} \left\vert \int_0^2(3th^2+6xh^2+2h^3) \ dt \right\vert &\le \int_0^2 \left\vert 3th^2+6xh^2+2h^3 \right\vert \ dt\\ &\le \Vert h \Vert_\infty^2 \int_0^2(3t+6 \Vert x \Vert_\infty+2)dt \end{aligned}$$ for $\Vert h \Vert_\infty \le 1$. And you can now conclude as the RHS of last inequality is $o(\Vert h \Vert_\infty)$.