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For the Weierstrass M-test proof of the Picard-Lindelöf theorem, the required bound isTo see this, use Picard iteration starting with $y_0(x)=b$ andIf $y_n$ takes its values in $[b-k,b+k]$, then $|y_{n+1}-b|\le M|x-a|\le Mh\le k$. Thus every Picard iterate remains in the rectangle where the hypotheses hold. Successive applications of the Lipschitz condition and integration give the factorial estimate for Picard iterationFor example, the $n=0$ estimate is $M|x-a|$, and integrating the $n$th estimate multiplied by $L$ proves the next one. The sum of the uniform bounds isfor every finite $Lh$, by the exponential series. Hence the telescoping series of Picard iterates converges uniformly by the Weierstrass M-test. The uniform limit stays in the closed rectangle, and the Lipschitz condition permits passage to the limit in the integral equation. The fundamental theorem of calculus then gives the desired solution of a differential equation on $[a-h,a+h]$.

With the standard constant starting Picard iterate, $y_0(x)=1$, the next two Picard iterates areand, for $x>0$,ThusThe restriction $x>0$ for $y_2$ is essential for the integral from $1$ to $x$: the preceding iterate $y_1$ vanishes at $0$, where the new integrand $s+1/s$ is undefined.

The difference of two successive Picard iterates factors asThe preceding bound and the kernel estimate therefore giveMoreover $D_1:=\Vert y_1-y_0{\Vert }_{\infty }\le B^2/C_1$. Summing the resulting geometric series, for $m>n$,Thus there is a uniform limit $y$ on $[0,\infty )$, continuous by the uniform limit theorem, with $\Vert y{\Vert }_{\infty }\le 2B$. The integral passes to this limit becauseTherefore $y$ satisfies the Volterra integral equation, and hence the initial value problem. This proves global existence by Global Picard iteration with a decaying kernel. The same estimate gives the useful error boundIf $b=0$, every iterate is zero and the convergence assertion is immediate.

For $s>t$, the coefficient bound gives $A(s)-A(t)\ge C_1(s-t)$. Thus for $0\le s\le t$ the positive kernel satisfiesThe starting Picard iterate is continuous and has supremum norm $M_0=|b|$. If the $n$th Picard iterate is bounded and continuous, its integral defines another continuous function, andTaking the supremum norm and applying mathematical induction proves boundedness of every iterate and

A Picard iterate is one of the successive approximate solutions of a differential equation obtained by repeatedly applying its integral solution operator. If $y^{\prime }=f(t,y)$ and $y(0)=b$, the usual operator is $(Ty)(t)=b+{\int }_0^{t}f(s,y(s))ds$. An integrating factor can produce a different, equivalent operator with better estimates, as in Global Picard iteration with a decaying kernel. The word iterate denotes an approximate function; Picard iteration denotes the procedure.