# Difficulty proving continuity of integral operator.

I am trying to prove that given $$g:I^{n+1}\subset \mathbb{R}^{n+1}\rightarrow \mathbb{R}^n$$ continuous, $(x_0,y_0)\in I^{n+1}$ and a sequence of functions defined iteratively as for each $f:\mathbb{R}\rightarrow \mathbb{R}^n$, $$f_0(x)=y_0\\ f_{n+1}(x)=\vec{y_0}+\int_{x_0}^xg(s,f_n(s))ds$$ is continuous on the compact k-cell $I^{n}$. I am not seeing an obvious way to bound the difference $$||\int_{x_0}^xg(s,f_n(s))ds-\int_{x_0}^yg(s,f_n(s))ds||$$ in terms of $||x-y||$. If it were a single variable integral this would be easy, but I am having a hard time trying to pluck out $||x-y||$ terms from anywhere.

Some context: I have (hopefully) shown uniform boundedness of the $f_n$ by arguing, for $L=||I^{n}||$ and $||g||_\infty\leq M$, $$||f_n(x)||_\infty=||\vec{y_0}+ \int_{x_0}^xg(s,f_n(s))\mathrm ds||_\infty\\ \stackrel{\text{triangle ineq.}}{\leq}||y_0||_\infty+ ||\int_{x_0}^xg(s,f_n(s))\mathrm ds||_\infty\\ \stackrel{\text{max of integrand times max volume}}{\leq} ||y_0||_\infty+ML$$ a constant.

This is in the context of proving a variation of the existence theorems for ODE's using Arzela-Ascoli.

edit: addressing comment: I may have left out some context because I was confused myself. The $g$ in this problem arises in the context of the system of differential equations $$f'(x)=g(x,f(x))\\ f(x_0)=y_0$$ for a continuous $g$ defined on an open interval $\mathcal{O}\subset \mathbb{R}^n\times \mathbb{R}\rightarrow \mathbb{R}^n$. I took $I^{n+1}$ to be some compact product of closed rectangles living inside $\mathcal{O}$, analogous to the picard theorem in 1 dimension.

The $f_n$ are approximating the differentiable $f:\mathbb{R}^n\rightarrow \mathbb{R}$.

• Can you clarify a couple points in your post? You state that $g$ is defined on $I^{n+1}$, but then you don't verify that your $f_n$ functions take values in $I^n$ (though you prove a bound, I don't see how to guarantee it's less than unity). Without proving this, how does $g(s,f_n(s))$ even make sense? Also, what do you mean by $L= \Vert I^n \Vert_\infty$? The $L^\infty$ norm is for functions, not sets. Finally, do you mean $\Vert g \Vert_\infty \le M$? Why do you have $x$ in there? – Glitch Jan 19 '17 at 4:38
• Yes, sorry, editing above. However, I am unclear on why the bound needs to be less than unity? To apply Arzela Ascoli I only need a uniform/constant bound. – qbert Jan 19 '17 at 4:47
• You never really define the set $I^n$. I was assuming $I = [-1,1]$. Is that not the case? If it is the case, then for $g(s,f(s))$ to make sense we have to have $(s,f(s)) \in I^{n+1}$. – Glitch Jan 19 '17 at 4:49
• @Glitch I have hopefully clarified in the above – qbert Jan 19 '17 at 4:54

As long as you know that $g(s,f_n(s))$ always makes sense (i.e. $(s,f(s))$ lies in the domain of the function $g$), and $g$ is bounded by $M$ on this set, then it's not too bad to get what you're after. Assume for the moment that $y \le x$. Then $$\int_{x_0}^x g(s,f_n(s)) ds - \int_{x_0}^y g(s,f_n(s)) ds = \int_{y}^x g(s,f_n(s)) ds$$ and so $$\left\Vert \int_{x_0}^x g(s,f_n(s)) ds - \int_{x_0}^y g(s,f_n(s)) ds \right\Vert \le \int_y^x \Vert g(s,f_n(s)) \Vert ds \le \int_y^x M ds = M |y-x|.$$ A similar argument proves the bound when $y \ge x$. Thus we have something better than continuity: we have Lipschitz continuity.
• Yes, the integrand (i.e. $g$) takes values in $\mathbb{R}^n$, but the bounds of integration are scalars. Moreover, for a vector integral of a function $F:[a,b] \to \mathbb{R}^n$ we can bound $\Vert \int_a^b F(s) ds \Vert \le \int_a^b \Vert F(s) \Vert ds$. – Glitch Jan 19 '17 at 5:02
• The last bit I knew. Thank you for bearing with me. Final question, if I want the norm of the product of intervals, I thought using $||\cdot||_\infty$ meant take the maximal element, or maximum of the absolute value of all the vector entries. How should I denote this? – qbert Jan 19 '17 at 5:04
• If I understand correctly you want notation for the supremal value of $\Vert x \Vert$ for $x \in I^{n+1}$. Is that right? If so, the closest thing I can think of is the "diameter of the set," which is the supremal distance between any two elements, but that's not quite what you want. Maybe just write $L = \sup_{x \in I^{n+1}} \Vert x \Vert$ ? – Glitch Jan 19 '17 at 5:08