# Proof Check: Mean value theorem and Lipschitz condition

for a function $$f(t,y)$$ to satisfy a lipchitz condition in the variable y on the set $$D = \{ (t,y) | a \leq t \leq b, -\infty < y < \infty\}$$, then there exists a constant $$\Lambda > 0$$ such that $$|f(t,y_1)-f(t,y_2)| < \Lambda |y_1 - y_2|$$ Apply the mean value theorem (for fixed t) to prove the following: a sufficient condition for Lipschitz condition is that $$f_{y}$$ exists and is bounded

Now, the course im currently studying is quite applied and so i believe there's quite a bit of leeway when it comes to stringent theory for example its implied that the set D is compact. in either case could i please get a proof check to make sure this is suitable

I believe A is sufficient for B suggests we're working with $$A \implies B$$ and thus:

let $$D = \{ (t,y) | a \leq t \leq b, -\infty < y < \infty\}$$ and define $$f:D \longrightarrow\mathbb{R}^{2}$$ to be reasonably well behaved such that $$\frac{\partial f}{\partial y}(t,y)$$ exists and is bounded by some real $$\Lambda > 0$$ for all $$(t,y) \in D$$

Fix t and let y vary then we have $$\frac{\partial f}{\partial y} = \lim_{h \longrightarrow 0} \frac{|f(t,y_2+h)-f(t,y_2)|}{|h|} \underset{y_1=y_2+h}{=} \quad\lim_{y_1 \longrightarrow y_2} \frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|} < \Lambda \underset{\text{Via Mean Value Theorem}}{\implies} |f(t,y_1) - f(t,y_2)| \leq \\ \frac{\partial f}{\partial y}(t,c) |y_1-y_2|$$ for some $$c \in D$$ then since $$\frac{\partial f}{\partial y}$$ is bounded by $$\Lambda$$ we have $$|f(t,y_1) - f(t,y_2)| \leq \frac{\partial f}{\partial y}(t,c) |y_1-y_2| < \Lambda |y_1-y_2|$$ Hence if the if function $$f$$ has a partial derivative in y with t fixed, as described above, then f satisfies the lipschitz condition as required.

I believe this should do it right? have i missed anything or are there any blaring holes? thanks in advance.

A final note: I have no idea where to post this other than Proof verification so if a friendly user with greater knoweldge than i could give a helping hand, i would be most grateful.

1. You write $$\frac{\partial f}{\partial y}$$ when you should be writting $$\left|\frac{\partial f}{\partial y}\right|$$ and you do not comment on the fact that you assume $$\frac{\partial f}{\partial y} > 0$$.
2. The main value theorem ensures that $$|f(t,y_1) - f(t,y_2)| = \left| \frac{\partial f}{\partial y}(t,c)\right||y_1-y_2|$$ for at least one $$c$$ between $$y_1$$ and $$y_2$$. You indicate that having a bounded partial derivative is a necessary condition for this to be true. It is not.
Let $$\Lambda \ge 0$$ be such that $$\left|\frac{\partial f}{\partial y}(t,y)\right| \leq \Lambda.$$ By assumption, we have $$\Lambda < \infty$$. We claim that $$f: D \rightarrow \mathbb{R}$$ is Lipschitz continuous with Lipschitz constant $$\Lambda \ge 0$$. To this end, let $$t \in [a,b]$$ and let $$y_1, y_2 \in \mathbb{R}$$ be given. We must show that $$|f(t,y_1)-f(t,y_2)| \leq \Lambda |y_1-y_2|.$$ Since $$y \rightarrow f(t,y)$$ is differentiable, the mean value theorem ensures that there exists at least one $$c$$ in the interval between $$y_1$$ and $$y_2$$ such that $$f(t,y_1) - f(t,y_2) = \frac{\partial f}{\partial y}(t,c)(y_1-y_2).$$ It follows that $$|f(t,y_1) - f(t,y_2)| = \left|\frac{\partial f}{\partial y}(t,c)\right||y_1-y_2| \leq \Lambda |y_1-y_2|.$$ This completes the proof.