Proof of Global Existence and Uniqueness for Lipschitz Ordinary Differential Equations in Integral Form

Let $(\mathcal{C},d_{\infty})$ be the nonempty complete metric space of functions $[a,b]\to\mathbb{R}^{k}$ with continuous components from Completeness of the Space of Continuous Vector-Valued Functions under the Supremum Metric, and write $|x|=d(x,0)$, so that $d(x,y)=|x-y|$ by claim 1 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals, whose componentwise inequalities we use freely.

The solution operator. For $h\in\mathcal{C}$ define

By hypothesis (i) the integrand has continuous components, so the componentwise Riemann integrals exist (Continuous Functions on a Closed Interval are Riemann Integrable) and each component of $Th$ is continuous on all of $[a,b]$, including the endpoints, by claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals. Thus $T:\mathcal{C}\to\mathcal{C}$, and a function $x$ is as in the conclusion exactly when it is a fixed point of $T$.

Basic integral bound. For $h,h'\in\mathcal{C}$ and $a\le t\le b$, claim 5 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals applied to $w(r)=F(r,h(r))-F(r,h'(r))$, together with hypothesis (ii) and monotonicity of the Riemann integral on continuous integrands (Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval with Linearity and Monotonicity of the Lebesgue Integral; the function $r\mapsto d(h(r),h'(r))$ is continuous by the componentwise estimates), gives, with $K=kL$,

Case $K=0$. Then $Th=Th'$ for all $h,h'$, so $T$ is a contraction (constant $\tfrac12$) on $(\mathcal{C},d_{\infty})$, and Contraction Mapping Theorem on a Nonempty Complete Metric Space gives a unique fixed point.

Case $K>0$: weighted metric. With the exponential function, define, for $h,h'\in\mathcal{C}$,

Since $\exp(-2K(b-a))\le\exp(-2K(t-a))\le1$ on $[a,b]$ (positivity and monotonicity from Basic Properties of the Exponential Function), every value defining $d_w$ lies between $\exp(-2K(b-a))$ times and $1$ times the corresponding value defining $d_\infty$; taking suprema,

$d_w$ is a metric on $\mathcal{C}$: nonnegativity and symmetry are clear; $d_w(h,h')=0$ forces $d_\infty(h,h')=0$ by the left inequality, hence $h=h'$; and multiplying the pointwise triangle inequality for $d$ by the positive weight $\exp(-2K(t-a))$ and passing to suprema gives the triangle inequality, exactly as in claim 1 of Completeness of the Space of Continuous Vector-Valued Functions under the Supremum Metric.

$(\mathcal{C},d_w)$ is complete. Let $(h_n)$ be Cauchy for $d_w$; given $\eta>0$, applying the Cauchy property with $\eta\exp(-2K(b-a))$ and using $d_\infty\le\exp(2K(b-a))\,d_w$ shows $(h_n)$ is Cauchy for $d_\infty$; by Completeness of the Space of Continuous Vector-Valued Functions under the Supremum Metric it converges to some $h\in\mathcal{C}$ for $d_\infty$; and $d_w(h_n,h)\le d_\infty(h_n,h)\to0$ shows convergence for $d_w$.

Contraction estimate. Fix $h,h'\in\mathcal{C}$ and $t\in[a,b]$. By the definition of $d_w$, $d(h(r),h'(r))\le\exp(2K(r-a))\,d_w(h,h')$ for every $r$, and

indeed $\exp(2K(r-a))=\exp(-2Ka)\exp(2Kr)$ by Basic Properties of the Exponential Function, the function $r\mapsto\exp(-2Ka)\exp(2Kr)/(2K)$ has derivative $\exp(2K(r-a))$ by Derivative of a Scaled Exponential Function and the constant-multiple rule of Sum and Product Rules for One-Dimensional Derivatives and Continuity, and Fundamental Theorem of Calculus, Part II in One Dimension evaluates the integral (degenerate $t=a$ by the convention of Mean-Square Riemann Integral of a Family of Random Variables). Combining with the basic integral bound and monotonicity,

Taking the supremum over $t$: $d_w(Th,Th')\le\tfrac12 d_w(h,h')$, so $T$ is a contraction of the nonempty complete metric space $(\mathcal{C},d_w)$.

By Contraction Mapping Theorem on a Nonempty Complete Metric Space, $T$ has exactly one fixed point $x\in\mathcal{C}$. Fixed points of $T$ are precisely the functions in the conclusion, so such an $x$ exists and is unique. $\blacksquare$