Find region for which F(x,y) = (x+y)^2 is Lipschitz in y

[jamesb1]

As the title says, I need to find such a region.

Taking any x, and any y1 and y2 I used the expression |F(x,y1) - F(x,y2)| and plugged in the function respectively for y1 and y2.

Now I have to find values for x and y such that the following condition (Lipschitz condition) is satisfied:

| 2x + (y1 + y2) | 0 (indeed after having simplified out the previous expression w.r.t the Lipschitz condition)

My initial idea was to find x for which y = 0 and then the same thing for y1 and y2. This method though is not enough since for x = 0 the region of y1 and y2 for which the condition is satisfied, will have to depend on y1 and y2 directly. It will be better to attain a region such that it will not depend on the variables (obviously). I hope I am not incorrect here.

I cannot seem to find a way to get this region and I would very much appreciate any insight given.

Thank you.

 

[pasmith]

As the title says, I need to find such a region.

Taking any x, and any y1 and y2 I used the expression |F(x,y1) - F(x,y2)| and plugged in the function respectively for y1 and y2.

Now I have to find values for x and y such that the following condition (Lipschitz condition) is satisfied:

| 2x + (y1 + y2) | 0 (indeed after having simplified out the previous expression w.r.t the Lipschitz condition)

My initial idea was to find x for which y = 0 and then the same thing for y1 and y2. This method though is not enough since for x = 0 the region of y1 and y2 for which the condition is satisfied, will have to depend on y1 and y2 directly. It will be better to attain a region such that it will not depend on the variables (obviously). I hope I am not incorrect here.

I cannot seem to find a way to get this region and I would very much appreciate any insight given.

Thank you.

For each $x \in \mathbb{R}$ and each $K > 0$ there exists an interval $$L_K(x) = \left[-\tfrac12K - x,\tfrac12K - x\right]$$ such that if $y_1 \in L_K(x)$ and $y_2 \in L_K(x)$ then $|F(x,y_1) - F(x,y_2)| \leq K|y_1 - y_2|$, so $f_x : y \mapsto F(x,y)$ is lipschitz with respect to $y$ with lipschitz constant $K$.

Note that you must fix $x$ and then determine the interval; if you want a region of the $(x,y)$-plane in which $F$ is lipschitz with respect to $y$ with constant $K$ then the condition is
$$|F(x_1,y_1) - F(x_2,y_2)| = |x_1^2 - x_2^2 + 2x_1y_1 - 2x_2y_2 + y_1^2 - y_2^2| \leq K|y_1 - y_2|.$$

 

[micromass]

Please post such questions in the homework forum in the future I'll move it to there now!

 

[jamesb1]

Shouldn't the interval not contain variables though, to have a definite region? I know you can choose any x and fix it, but it seems indefinite to me.

x should remain fixed instead of having x1 and x2, no? I am unsure how to attain values for both x and y such that the following inequality is satisfied:

| 2x + (y1 + y2) | <= K