Picard-Lindelöf Theorem

Let $\varepsilon > 0$ be arbitrary and $(f_n) \subseteq C(K,E)$ be a Cauchy sequence, then there exists $N \in \mathbb{N}$ such that $\|f_n - f_m\|_{\infty} < \varepsilon$ whenever $n,m > N$. Fix $x \in K$, then we have

Then $(f_n(x))$ is Cauchy in $E$. Since $E$ is complete, there exists $f(x) \in E$ such that $f_n(x) \to f(x)$. We claim that $f_n \to f$ uniformly. Fix $n > N$ and $x \in K$, then we have

Then $f_n \to f$ in $E$. In addition, taking $m \to \infty$ we obtain $\|f_n - f\|_\infty \leq \varepsilon$ for all $n \geq N$. Thus $f_n \to f$ in the sup norm. To prove $f \in C(K,E)$, fix $x_0 \in K$, we choose $N \in \mathbb{N}$ large enough such that

Since $f_n$ is pointwise continuous, then there exists $\delta > 0$ such that $\|f_n(x) - f_n(x_0)\| < \frac{\varepsilon}{3}$ whenever $d(x,x_0) < \delta$. Then we obtain

Since $x_0 \in K$ was arbitrary, then $f \in C(K,E)$, which implies that $C(K,E)$ with the supremum norm is a Banach space.

Let $x_0 \in X$ be arbitrary and let $(x_n)$ be the sequence defined by $x_n = T(x_{n-1})$ for all $n \geq 1$. We first establish the estimate

This follows by induction: the base case $n = 1$ is immediate. Assuming it holds for $n$, then

Using this and the triangle inequality, for all $m > n$ we obtain

Let $\varepsilon > 0$ be arbitrary, then one can choose $N \in \mathbb{N}$ large enough such that $\frac{q^n}{1-q}d(x_1,x_0) < \varepsilon$. Thus $(x_n)$ is a Cauchy sequence, and since $X$ is complete, then $x_n$ converges to some $x^* \in X$ and we obtain

where the last equality follows from continuity of $T$. Thus $x^*$ is a fixed point in $X$. To prove the uniqueness, suppose $p_1 \neq p_2$ are two fixed points in $X$, then we have

which is a contradiction, as desired.

Let $v(t) = C + L\int_{t_0}^t u(s)\,ds$, then we have $u(t) \leq v(t)$ and $v'(t) = Lu(t) \leq Lv(t)$. Consider $w(t) = e^{-L(t-t_0)}v(t)$, differentiating $w$ yields

Thus $w(t) \leq w(t_0) = v(t_0) = C$, we obtain

Hence we are done.

Let $a, r \in \mathbb{R}$ such that $r > 0$ and $W = [t_0 - a, t_0 + a] \times \overline{B(x_0,r)} \subseteq U$. Since $W$ is closed and bounded, it is compact. Then $F$ is bounded on $W$.

Let $M = \sup_{(t,x) \in W}\|F(t,x)\|$, $b = r$, $h = \min\left\{\frac{b}{M}, \frac{1}{2L}, a\right\}$ and $I = [t_0 - h, t_0 + h]$. We define the metric space $(X, \|\cdot\|_{\infty})$

Since $I$ is compact and $\mathbb{R}^n$ is complete, the above lemma gives that $C(I,\mathbb{R}^n)$ is Banach. Since $X \subseteq C(I,\mathbb{R}^n)$ is closed, then $X$ is also a Banach space. Integrating both sides of the second equation of $(1)$ yields

Let $T: X \to C(I, \mathbb{R}^n)$ be the operator defined by $T[x](t) = x_0 + \int_{t_0}^t F(s, x(s))\,ds$. We claim that $T$ is invariant. Indeed, we have

Thus $T[x] \in X$. Furthermore, we aim to prove that $T$ is a contraction. Since $W$ is compact, then there exists $L > 0$ such that $F$ is Lipschitz with constant $L$ on $W$. Consider the following estimate

Taking the supremum over $t \in I$ yields $\|Tx - Ty\|_\infty \leq \frac{1}{2}\|x - y\|_\infty$. Thus $T$ is a contraction. Applying the Banach Fixed Point theorem, then there exists a unique $x^* \in X$ such that $T[x^*] = x^*$, or equivalently,

Then $F(s, x^*(s))$ is continuous, the fundamental theorem of calculus implies that

Hence $x^*$ is a solution in $C^1(I,\mathbb{R}^n)$ of $(1)$. To prove the uniqueness, suppose $x_1, x_2 \in C^1(I,\mathbb{R}^n)$ are solutions of $(1)$. Since both satisfy the integral equation, using the similar estimate we obtain

If $t < t_0$, the integral $\int_{t_0}^t = -\int_t^{t_0}$ is nonpositive while the left side is nonnegative, so both sides must vanish, which implies $\|x_1(t) - x_2(t)\| = 0$, i.e. $x_1(t) = x_2(t)$. If $t \geq t_0$, apply the Grönwall inequality for the function $u(t) = \|x_1(t) - x_2(t)\|$ with $C = 0$, we obtain $u(t) \leq 0$ for all $t \geq t_0$, thus $u(t) = 0$. Therefore $x_1(t) = x_2(t)$ for all $t \in I$, it follows that the solution $x^*$ in $C^1(I,\mathbb{R}^n)$ is unique.