Question: Suppose we are given $f(x,t)$ a uniformly lipschitz continuous function in $x$.

That is to say for any $x, \hat{x}$ in $\mathbb{R}$ we have $|f(x,t) - f(\hat{x},t)| \leq L|x-\hat{x}|$.

I want to show that $|f(x,t)| \leq L(1+|x|)$.

Context: I am reading Evan's Introduction to SDE's, and I am currently in the proof regarding the existence and uniqueness of solutions to SDEs. The original problem itself is multidimensional and I will post the statement below for completeness.


Suppose that $\textbf{b}: \mathbb{R}^n \times [0,T] \to \mathbb{R}^n$ and $\textbf{B}: \mathbb{R}^n\times [0,T] \to \mathbb{R}^{m\times n}$ are continuous and satisfy the following conditions.

  1. For some constant $L$ and all $0 \leq t \leq T$, $x, \hat{x} \in \mathbb{R}^n$ we have \begin{equation*} \begin{split} |\textbf{b}(\textbf{x}, t) - \textbf{b}(\hat{\textbf{x}}, t)| &\preceq L|\textbf{x}-\hat{\textbf{x}}| \\ |\textbf{B}(\textbf{x}, t) - \textbf{B}(\hat{\textbf{x}}, t)| &\preceq L|\textbf{x}-\hat{\textbf{x}}| \end{split} \end{equation*}
  2. For the same constant $L$ (as above) and all $0 \leq t \leq T$, $x \in \mathbb{R}^n$ we have \begin{equation*} \begin{split} |\textbf{b}(\textbf{x}, t)| &\preceq L(1+|\textbf{x}|) \\ |\textbf{B}(\textbf{x}, t)| &\preceq L(1+|\textbf{x}|) \end{split} \end{equation*} (... The rest of the theorem statement is irrelevant to this question)

And after stating the theorem, the author says "It is possible to show that $(1) \implies (2)$" directly. On the onset this sounds believable, but somehow, I am getting stuck.

My attempt: WLOG, we let $n=1$, and try only to solve the first part with $\textbf{b}$ as the other part is done similarly. Notice that letting $\hat{x} = 0$ we have $$|f(x,t)|-|f(0,t)| \leq |f(x,t) - f(0,t)| \leq L|x|$$ This tells us that $$|f(x,t)| \leq |f(0,t)| + L|x|$$

At this point, I am forced to somehow show $|f(0,t)| \leq L$, but since no additional assumptions on $f$ is given, I find this hard to believe. Intuitively, the lipschitz condition only bounds the derivative (wrt $x$) of $f$ by $L$, but the function value itself can be anything it wants. However, if the statement I am trying to prove is true, then indeed $|f(0,t)| \leq L$. So, some miraculous force is stopping $|f(0,t)|$ from going beyond $L$; any help in helping me figure out what that might be is very much appreciated!

Resolution: The comment from @Shalop was a great first step, but the details had to be ironed out; I am doing it here for my own references, and just in case someone else needed help with this stuff. Let $T>1$. The example $f(x,t) = t^2 + x$ seems to work on the onset because $|f(x,t) - f(\hat{x},t)| \leq |x - \hat{x}|$, so $L=1$ and $|f(0,t)| = t^2$. However, $L$ doesn't necessarily have to be $1$; in this case if we let $L = T^2$; then we still have

$$|f(x,t) - f(\hat{x},t)| \leq T^2|x-\hat{x}|$$ and at the same time $$|f(0,t)| \leq L$$ giving us what we need.

However, motivated by this first approach, I wrote a different proof, which I think works.

Let $f$ satisfy the given condition. By hypothesis, we are given some $L$ such that $$|f(x,t) - f(\hat{x},t)| \leq L|x - \hat{x}| \quad \forall{x, \hat{x} \in \mathbb{R}, t \in [0,T]}$$ Now consider $g(x,t) = f(x,t)+L+1-f(0,t)$. Notice that for the same $L$, $g$ also satisfies $$|g(x,t) - g(\hat{x},t)| \leq L|x - \hat{x}|$$ However, if the hypothesis did in fact imply the conclusion, this would mean $|g(0,t)| \leq L$; but by construction, we know that $|g(0,t)| = |f(0,t)+L+1-f(0,t)| = L+1$, which is a contradiction!

So, we have found a $g$ that satisfies the hypothesis, but fails to satisfy $g(0,t) \leq L$.

  • 3
    $\begingroup$ Seems to me like $f(x,t) = t^2+x$ satisfies your condition. However it does not satisfy the desired bound. $\endgroup$
    – shalop
    Mar 1, 2021 at 3:21
  • $\begingroup$ Oh yeah! This does indeed work if $T > 1$ (since $L=1$ here), thank you for the nice counter-example. (I thought I was going crazy trying to prove this!) $\endgroup$ Mar 1, 2021 at 3:34
  • $\begingroup$ @shalop, This was a great first step, but as I was writing the proof down formally, I noticed that this didn't quite work fully. So motivated by this, I wrote a different proof that I think works; and I've edited the question to add the last \textbf{Resolution} bit for completion. (Just thought I would let you know haha) $\endgroup$ Mar 2, 2021 at 0:19
  • $\begingroup$ I think it’s clear that (1) does imply (2) if you allow the constant in (2) to be a bit larger than the one in (1), depending on f. I imagine that this is what the author probably meant. My example really only works if you allow for unbounded time intervals. $\endgroup$
    – shalop
    Mar 2, 2021 at 4:41
  • $\begingroup$ Yes, if we have different constants for (1) and (2), the result is straightforward; and I am starting to think that might be what the author could have meant. My problem was only with the fact that given a constant in (1), I am unable to a-priori prove (2) with the same constant. If the constant in (1) had the freedom to be changed based on the bounds for (2), the result is more or less direct like you mentioned $\endgroup$ Mar 2, 2021 at 5:38


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