# Continuous functions, Lifchitz's condition...

• Jun 14th 2011, 04:13 PM
Also sprach Zarathustra
Continuous functions, Lifchitz's condition...
Let $u=f(x,y)$ defined in $D$, continuous by $x$ and holds Lifchitz's condition for $y$, in other words: $|f(x,y_1)-f(x,y_2)|\leq A|y_1-y_2|$ where $(x,y_1),(x,y_2)$ are points in $D$ and $A$ is constant. Prove $f(x,y)$ continuous in $D$.

I know that $f(x,y)$ is continuous by $y$ (from Lifchitz's condition) and it's given that $f(x,y)$ is continuous by $x$, but now what?

Thank you!
• Jun 14th 2011, 05:04 PM
Drexel28
Re: Continuous functions, Lifchitz's condition...
Quote:

Originally Posted by Also sprach Zarathustra
Let $u=f(x,y)$ defined in $D$, continuous by $x$ and holds Lifchitz's condition for $y$, in other words: $|f(x,y_1)-f(x,y_2)|\leq A|y_1-y_2|$ where $(x,y_1),(x,y_2)$ are points in $D$ and $A$ is constant. Prove $f(x,y)$ continuous in $D$.

I know that $f(x,y)$ is continuous by $y$ (from Lifchitz's condition) and it's given that $f(x,y)$ is continuous by $x$, but now what?

Thank you!

Note that for any [tex](x_0,y_0)\in D[tex] you have that $\left\|f(x,y)-f(x_0,y_0)\|\leqslant \|f(x,y_0)-f(x,y)\|+\|f(x_0,y_0)-f(x,y_0)\|$ now if you make a neighborhood small enough to make the first less than epsilon over two by continuity and the second epsilon/(2A) by the Lipschitz condition you're done.
• Jun 14th 2011, 05:05 PM
Also sprach Zarathustra
Re: Continuous functions, Lifchitz's condition...
Great!!! Thank you!
• Jun 14th 2011, 06:04 PM
Also sprach Zarathustra
Re: Continuous functions, Lifchitz's condition...
Quote:

Originally Posted by Drexel28
Note that for any [tex](x_0,y_0)\in D[tex] you have that $\left\|f(x,y)-f(x_0,y_0)\|\leqslant \|f(x,y_0)-f(x,y)\|+\|f(x_0,y_0)-f(x,y_0)\|$ now if you make a neighborhood small enough to make the first less than epsilon over two by continuity and the second epsilon/(2A) by the Lipschitz condition you're done.

I don't understanding something...

How is that second term $||f(x_0,y_0)-f(x,y_0) ||$ is less than epsilon/(2A) by the Lipschitz condition?( $|f(x_0,y_0)-f(x,y_0) |$\leq A|y_0-y_0|=0 )

Hmmm...(Sleepy)
• Jun 14th 2011, 07:43 PM
Drexel28
Re: Continuous functions, Lifchitz's condition...
Quote:

Originally Posted by Also sprach Zarathustra
I don't understanding something...

How is that second term $||f(x_0,y_0)-f(x,y_0) ||$ is less than epsilon/(2A) by the Lipschitz condition?( $|f(x_0,y_0)-f(x,y_0) |$\leq A|y_0-y_0|=0 )

Hmmm...(Sleepy)

Let $(x_0,y_0)\in D$ be arbitrary. By continuity in the $x$ variable there exists some $\gamma>0$ such that if $|x-x_0|<\gamma$ we have that $\displaystyle f(x,y)-f(x_0,y)|<\frac{\varepsilon}{2}$. Let $\displaystyle \delta=\min\left\{\gamma,\frac{\varepsilon}{2A}\ri ght\}$. Let then $(x,y)\in B_\delta(x_0,y_0)$. We know that $|f(x_0,y_0)-f(x,y)|\leqslant |f(x,y_0)-f(x_0,y_0)|+|f(x,y_0)-f(x,y)|$. Now, since $|x-x_0|\leqslant \|(x,y)-(x_0,y_0)\|<\gamma$ we have that this first term is less than $\displaystyle \frac{\varepsilon}{2}$ and this second term is, by Lipschitz, less than $\displaystyle A|y-y_0|\leqslant A\|(x,y)-(x_0,y_0)\|.