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I want to ask one general question, and after that I would like to know if my method is correct (for determining whether a function is Lipschitz with respect to y)

Is the following statement true?

Suppose D is a convex domain, the function $f(x,y)$ is continuous and $f_y$ is continuous in D. Then $\sup|f_y|\le L \iff |f(x,y_1)-f(x,y_2)| \le L|y_1-y_2|$

(In other words: Iff $f$ and $f_y$ are continous in a convex domain D, then f is lipschitz, with Lipschitz constant L as defined in my theorem)

A few examples (supposing $|x| \le 1$)

$$f_1(x,y)=x^4e^{-y^2}$$ $$f_2(x,y)=\frac{1}{1+y^2} $$ $$f_3(x,y)=x^2+y^2 $$ $$f_4(x,y)= |y| \text{ if |y|}\le1, f_4=1 \text{ if } |y|\ge 1 $$

My work would be:

$f_1$ is continuous, $f_y=-2yx^4e^{-y^2} \le -2ye^{-y^2}$ is continous everywhere. $\sup|f_y|=\sqrt{2/e}$, so $f_1$ satisfies Lipschitz constant $L=\sqrt{2/e}$

$f_2$ is continuous, $\sup|f_y|=sup|\frac{-2y}{(1+y^2)^2}|=\frac{3\sqrt{3}}{8} $

$f_3$ is continous, $f_y=2y$, $\sup|2y|$ does not exists, because y is not bounded. Therefore $f(x,y)$ does not satisfy the Lipschitz condition. Another way to see this is that, for $y_2=0$ if $|f(x,y_1)-f(x,y_2)| =| y_1^2| \le L|y_1 - y_2|=L|y_1| \implies \frac{y_1^2}{y_1}=y_1\le L$, which is not true because y is not bounded.

I have some trouble understanding the Lipschitz condition for $f_4$. Maybe somebody can help me analysing this function?

Thank you in advance...

(Please give me hints, tips and tricks, because I want to do it all myself).

share|cite|improve this question
Hint: Use the Mean Value Theorem in functions $y\mapsto \sup_{x} f(x,y)$. – MathOverview Dec 9 '12 at 16:09
This is a hint for which problem? What is $\sup_x$ – Applied mathematician Dec 9 '12 at 16:23
The Lipschitz condition is on $f$, not on $f_y$. $-$ What it is all about is the following: A $\sup$ condition on $f_y$ is equivalent with a Lipschitz condition on $f$. – Christian Blatter Dec 9 '12 at 16:26
bump.................... – Applied mathematician Dec 11 '12 at 16:33

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