= Solution
{c}
For the <Weierstrass M-test> proof of the <Picard-Lindelöf theorem>, the required bound is
$$
\boxed{Mh\leq k,\qquad\text{with no additional restriction on }Lh.}
$$
To see this, use <Picard iteration> starting with $y_0(x)=b$ and
$$
y_{n+1}(x)=b+\int_a^x f(s,y_n(s))\,ds.
$$
If $y_n$ takes its values in $[b-k,b+k]$, then $|y_{n+1}-b|\leq M|x-a|\leq Mh\leq k$. 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 iteration>
$$
|y_{n+1}(x)-y_n(x)|\leq\frac{ML^n|x-a|^{n+1}}{(n+1)!}.
$$
For 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 is
$$
\sum_{n=0}^{\infty}\frac{ML^nh^{n+1}}{(n+1)!}<\infty
$$
for 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 $[a-h,a+h]$.
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