Let $f:V\rightarrow \mathbb{R}^n$ be locally Lipschitz ($V$ is a subset of $\mathbb{R}\times\mathbb{R}^m\times \mathbb{R}^n$). Suppose we have a function $x:[t_0,\beta[\times W\rightarrow \mathbb{R}^n$ differentiable in the first argument ($W$ is an open subset of $\mathbb{R}^m$, $\beta$ is finite) such that for every $(t,\overrightarrow{\alpha})\in [t_0,\beta[\times W$ we have:

$$(t,\overrightarrow{\alpha},x(t,\overrightarrow{\alpha}))\in V$$


Here $x_1(t,\overrightarrow{\alpha})$ means partial derivative with respect to first argument.

It is also given that the function $g:W\rightarrow \mathbb{R}^n$ given by $g(\overrightarrow{\alpha})=x(t_0,\overrightarrow{\alpha})$ is locally Lipschitz.

Question: Does it follow that the function $x:[t_0,\beta[\times W\rightarrow\mathbb{R}^n$ is continuous ?

I can only prove the conclusion if the hypotheses are strengthened to $f,g$ Lipschitz instead of just merely locally Lipschitz.I would still like to know the answer in the locally Lipschitz case.

Thank you a lot.


The setup may be rewritten in the following form (avoiding special treatment of the parameter):

$\;\;y=(\alpha,x)$, $\; \; y_0=(\alpha,g(\alpha))$, $\; \;v(t,y) = (0,f(t,\alpha,x))$ and the ode: $$ \dot{y} = v(t,y), \ \ y(t_0) = y_0 $$ Now, if $v$ is locally Lipschitz in $y$ then for each $y_0$ there is a maximal solution defined (we apparently only look at $t\geq t_0$) some interval of time $[t_0,\tau)$ where $\tau=\tau(y_0)$ is a lower semi-continuous function of the initial condition $y_0$.

For every $t_0<b<\tau$ there is a neighborhood $U=U_{t_0,b}(y_0)$ (which may be very small) so that a maximal solution $y=\phi^t(t_0,y_1)$ exists for all $y_1\in U$, $t\in [t_0,b]$ and in addition $y_1\in U\mapsto \phi^t(t_0,y_1)$ is $L$-Lipschitz for all $t\in [t_0,b]$ and some $L=L(U)<+\infty$.

[The proof uses the local result which it seems you have understood how to get, and then uses compactness of $[t_0,b]$ to show that the Lipschitz dependency pops out from composing a finite number of local solutions]

Your hypothesis is that given the initial condition $y_0=y_0(\alpha) =(\alpha,g(\alpha))$, with $\alpha \in W$ the maximal solution exists up to, but not including the time $\beta$. Then for any $t_0<b<\beta$ the (composed) flow: $$ (t, \alpha)\in [t_0,b] \times W \mapsto \phi^t(t_0, (\alpha,g(\alpha)))$$ is locally Lipschitz in $\alpha$. Taking the limit $b\rightarrow \beta^-$ you infer that the map is continuous on $[t_0,\beta[\times W$ (but may fail to be locally Lipschitz on this set).


Let's fix $t_0 = 0$. We will consider the case when $f$ doesn't depend on $t$. I find it easier to prove the more general (well-known) statement

Let $f : V \rightarrow \mathbb{R}^n$ be locally Lipschitz ($V$ is open) . Define $\displaystyle D := \bigcup_{x \in V} I_x \ \times \{x\} \subseteq \mathbb{R} \times V $, where $I_x$ is the unique maximal interval of the unique solution $y(\cdot,x):I_x \rightarrow \mathbb{R}^n$ of $$ y' = f(y) $$ Then $D$ is open and $y:D \rightarrow \mathbb{R}^n$ is continuous, we call this the local flow of $f$

Your claim follows by letting $Y = (\alpha,x):D \subseteq V \rightarrow \mathbb{R}^m \times \mathbb{R}^n$ be the local flow of $F=(0,f)$. In other-words, for all $(t,\alpha_0,x_0) \in D$.

$$ \alpha'(t,\alpha_0,x_0) = 0 \quad ; \quad x'(t,\alpha_0,x_0) = f(\alpha_0,x(t,\alpha_0,x_0))$$

with $\alpha(0,\alpha_0,x_0) = \alpha_0$ and $x(0,\alpha_0,x_0)=x_0$.

Now your assumption:

" Suppose we have a function $x:[0=t_0,\beta) \times W\rightarrow \mathbb{R}^n$ differentiable .... "

Can be restated by saying that $[0,\beta) \times \mathrm{graph}(g) \subseteq D$. By the theorem above, we conclude that the following composition is continuous;

$$[0,\beta) \times W \hookrightarrow [0,\beta) \times \mathrm{graph}(g) \subseteq D \xrightarrow{Y} \mathbb{R}^m \times \mathbb{R}^n \xrightarrow{\pi_2} \mathbb{R}^n $$

where the first map is the obvious one, and $\pi_2$ denote the projection on the "second" argument. This is clearly "your" map.


I will argue that yes, if that $x:[t_0,\beta[\times W\to\mathbb R^n$ exists then it is continuous.

Let's prove sequential continuity: fix $(t_n,\alpha_n)$ converging to $(t^*,\alpha_0)$ (all in $[t_0,\beta[\times W$). We need to show $x(t_n,\alpha_n)\to x(t^*,\alpha_0).$ We will construct a neighbourhood of $(t^*,\alpha_0)$ on which $x$ is continuous, and since some tail of the sequence $(t_n,\alpha_n)$ is in this neighbourhood, this will give the necessary convergence. Set $T=\tfrac12(t^*+\beta),$ so $t_0\leq t^*<T<\beta.$

By Lemma 1 below there exist $A,X,L>0$ such that whenever $t,\alpha$ satisfy:

  • $0\leq t\leq T$ and
  • $\|\alpha-\alpha_0\|\leq A$ and
  • $\|x(t,\alpha)-x(t,\alpha_0)\|\leq X$


$$\|x_1(t,\alpha)-x_1(t,\alpha_0)\|\leq L(\|\alpha-\alpha_0\|+\|x(t,\alpha)-x(t,\alpha_0)\|).\tag{*}$$

If necessary, shrink $A$ further to ensure that $g$ is Lipschitz on the region $\|\alpha-\alpha_0\|\leq A.$ This is possible by the local Lipschitz property of $g.$

By continuity of $g$, we can shrink $A$ if necessary to also ensure that whenever $\|\alpha-\alpha_0\|\leq A$ we have

$$\|\alpha-\alpha_0\|, \|g(\alpha)-g(\alpha_0)\|\leq \tfrac 1 2 X e^{-LT}.$$

We will now show:

$$\|x(t,\alpha)-x(t,\alpha_0)\|\leq(\|\alpha-\alpha_0\|+\|g(\alpha)-g(\alpha_0)\|)e^{Lt}\tag{G}$$ for all $0\leq t\leq T$ and $\|\alpha-\alpha_0\|\leq A.$ Note that the right-hand-side is at most $X.$

Suppose not. Fix some $\alpha$ for which (G) fails. There are two cases:

  • (G) fails very badly: $\|x(t,\alpha)-x(t,\alpha_0)\|\geq X$ for some $0\leq t<T$ and $\|\alpha-\alpha_0\|\leq A$
  • (G) fails mildly: $\|x(t,\alpha)-x(t,\alpha_0)\|\leq X$ for some $0\leq t\leq T$ and $\|\alpha-\alpha_0\|\leq A$

For the very bad case, we can take $t$ to be infimal such that $\|x(t',\alpha)-x(t',\alpha_0)\|\geq X,$ for some fixed $\alpha.$ In the mild case, just pick any $(t,\alpha)$ such that (G) fails. We now apply Gronwall's inequality to (*); for both the very bad and mild cases, we have arranged that (*) applies for all smaller $t.$ Gronwall gives:

\begin{align*} \|x(t,\alpha)-x(t,\alpha_0)\| &\leq \|\alpha-\alpha_0\|+\|x(t,\alpha)-x(t,\alpha_0)\|\\ &\leq (\|\alpha-\alpha_0\|+\|g(\alpha)-g(\alpha_0)\|)e^{Lt} \end{align*} as required. This proves (G).

This means that solutions starting close to $\alpha_0$ stay in the set $\|x(t,\alpha)-x(t,\alpha_0)\|\leq X$. This means they obey an ODE with a global Lipschitz condition on $[0,T]$. We can therefore apply Lemma 2 below to get a neighbourhood of $(t^*,\alpha_0)$ on which $x$ is continuous.

Lemma 1

For this argument we need that there is a neighbourhood of $C=\{(t,\alpha_0,x(t,\alpha_0))\mid 0\leq t\leq T\}$ on which $f$ is Lipschitz. Note that $C$ is compact - it's the graph of the continuous function $x|_{[0,T]\times \{\alpha_0\}}.$ So this is a special case of the more general statement:

For all metric spaces $X,Y$, compact subsets $C\subseteq X$, and locally Lipschitz functions $f:X\to Y$, there exists a neighbourhood of $C$ on which $f$ is Lipschitz.

The proof is to suppose otherwise. Then for each $n\geq 1$ there are distinct points $x_n,x'_n$ in the $1/n$-neighbourhood $B_X(C,1/n)$ with $d_Y(f(x_n),f(x'_n))\geq n d_X(x_n,x'_n) $. By compactness of $C$ we can restrict to a subsequence $n_k\to\infty$ such that $x_{n_k}$ and $x'_{n_k}$ converge to some $x\in C$. Since $f$ is locally Lipschitz, there exist $L,\epsilon>0$ such that $f$ is $L$-Lipschitz on $B(x,\epsilon)$. But for sufficiently large $k$ we have $x_{n_k},x'_{n_k}\in B(x,\epsilon)$ and $n_k>L$, which contradicts the choice of $x_{n_k},x'_{n_k}.$

Lemma 2

If $f$ is Lipschitz on $V'\subseteq V$, and $g$ is Lipschitz on $W'\subseteq W$, and $(t,\alpha,x(t,\alpha))\in V'$ for all $(t,\alpha)\in[0,T]\times W'$, then $x$ is continuous on $[t_0,T)\times W'.$

This is stated as known in the original question.

  • $\begingroup$ Hi thanks for your post. How do we know that there exist $A,X,L$ with the properties you described ? $\endgroup$ – Amr Oct 9 '17 at 1:23
  • $\begingroup$ @Amr: I have added details for this point $\endgroup$ – Dap Oct 9 '17 at 8:48
  • $\begingroup$ Thank you for your edit $\endgroup$ – Amr Oct 10 '17 at 23:27
  • $\begingroup$ Thank you for your edit.I actually don't understand your argument. You start with assuming that $x (t,alpha) $ is within distance $X $ from $x (t,\alpha_0)$ (in your $A,X,L $ hypothesis) then you use grownwalls inequality to deduce again that $x (t,alpha) $ is within distance $X $ from $x (t,\alpha_0)$ which seems like an unnecessary loop for me $\endgroup$ – Amr Oct 10 '17 at 23:33
  • $\begingroup$ I fail to see why your argument shows that $x $ is continuous, may be I m missing something too obvious. I would like to see an argument that ends like this: $\endgroup$ – Amr Oct 10 '17 at 23:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.