2
$\begingroup$

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 $f:\mathbb{R}^n\rightarrow \mathbb{R}$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Glitch Jan 19 '17 at 4:38
  • $\begingroup$ 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. $\endgroup$ – qbert Jan 19 '17 at 4:47
  • $\begingroup$ 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 $(s,f(s)) \in I^{n+1}$. $\endgroup$ – Glitch Jan 19 '17 at 4:49
  • $\begingroup$ @Glitch I have hopefully clarified in the above $\endgroup$ – qbert Jan 19 '17 at 4:54
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ ah! I didn't know this was justified for vector valued integrals, but I think this confusion came from my initial thought that the variables in the limits were vectors. Just to clarify, we are integrating a vector valued function along a curve and thus can split things up as you did? $\endgroup$ – qbert Jan 19 '17 at 4:59
  • $\begingroup$ 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$. $\endgroup$ – Glitch Jan 19 '17 at 5:02
  • $\begingroup$ 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? $\endgroup$ – qbert Jan 19 '17 at 5:04
  • $\begingroup$ 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 $L = \sup_{x \in I^{n+1}} \Vert x \Vert$ ? $\endgroup$ – Glitch Jan 19 '17 at 5:08
  • $\begingroup$ ah ok, I will. Thanks again! $\endgroup$ – qbert Jan 19 '17 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.