Proof Check: Mean value theorem and Lipschitz condition

numerical methodssolution-verification

I've been asked the following:

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.

Best Answer

What you write is literally wrong and a sufficiently strict instructor would not accept this.

  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.

Do not be afraid of using many words to support your reasoning. I shall illustrate.

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.

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