# The Unapologetic Mathematician

## The Picard Iteration Converges

Now that we’ve defined the Picard iteration, we have a sequence of functions $v_i:J\to B_\rho$ from a closed neighborhood of $0\in\mathbb{R}$ to a closed neighborhood of $a\in\mathbb{R}^n$. Recall that we defined $M$ to be an upper bound of $\lVert F\rVert$ on $B_\rho$, $K$ to be a Lipschitz constant for $F$ on $B_\rho$, $c$ less than both $\frac{\rho}{M}$ and $\frac{1}{K}$, and $J=[-c,c]$.

Specifically, we’ll show that the sequence converges in the supremum norm on $J$. That is, we’ll show that there is some $v:J\to B_\rho$ so that the maximum of the difference $\lVert v_k(t)-v(t)\rVert$ for $t\in J$ decreases to zero as $i$ increases. And we’ll do this by showing that the individual functions $v_i$ and $v_j$ get closer and closer in the supremum norm. Then they’ll form a Cauchy sequence, which we know must converge because the metric space defined by the supremum norm is complete, as are all the $L^p$ spaces.

Anyway, let $L=\lVert v_1-v_0\rVert_\infty$ be exactly the supremum norm of the difference between the first two functions in the sequence. I say that $\lVert v_{i+1}-v_i\rVert_\infty\leq(cK)^iL$. Indeed, we calculate inductively

\displaystyle\begin{aligned}\lVert v_{i+1}(t)-v_i(t)\rVert&\leq\int\limits_0^t\lVert F(v_i(s))-F(v_{i-1}(s))\rVert\,ds\\&\leq K\int\limits_0^t\lVert v_i(s)-v_{i-1}(s)\rVert\,ds\\&\leq K\int\limits_0^t(cK)^{i-1}L\,ds\\&\leq(cK)(cK)^{i-1}L\\&=(cK)^iL\end{aligned}

Now we can bound the distance between any two functions in the sequence. If $i are two indices we calculate:

\displaystyle\begin{aligned}\lVert v_j-v_i\rVert_\infty&=\left\lVert\sum\limits_{k=i}^{j-1}v_{k+1}-v_k\right\rVert_\infty\\&\leq\sum\limits_{k=i}^{j-1}\lVert v_{k+1}-v_k\rVert_\infty\\&\leq\sum\limits_{k=i}^{j-1}(cK)^kL\end{aligned}

But this is a chunk of a geometric series; since $cK<1$, the series must converge, and so we can make this sum as small as we please by choosing $i$ and $j$ large enough.

This then tells us that our sequence of functions is $L^\infty$-Cauchy, and thus $L^\infty$-convergent, which implies uniform pointwise convergence. The uniformity is important because it means that we can exchange integration with the limiting process. That is,

$\displaystyle\lim\limits_{k\to\infty}\int_0^tv_k(s)\,ds=\int\limits_0^t\lim\limits_{k\to\infty}v_k(s)\,ds=\int\limits_0^tv(s)\,ds$

$\displaystyle v_{k+1}(t)=a+\int\limits_0^tF(v_k(s))\,ds$

and take the limit of both sides

\displaystyle\begin{aligned}v(t)&=\lim\limits_{k\to\infty}v_{k+1}(t)\\&=\lim\limits_{k\to\infty}\left(a+\int\limits_0^tF(v_k(s))\,ds\right)\\&=a+\int\limits_0^t\lim\limits_{k\to\infty}F(v_k(s))\,ds\\&=a+\int\limits_0^tF(v(s))\,ds\end{aligned}

where we have used the continuity of $F$. This shows that the limiting function $v$ does indeed satisfy the integral equation, and thus the original initial value problem.