Lipschitz continuity of $f(x,y)=4x^2+xy-\frac{1}{y-1}$ on an open set $U \subset \mathbb{R} \times (\mathbb{R} \setminus \lbrace 1 \rbrace)$

Problem: Find an open set $U \subset \mathbb{R} \times (\mathbb{R} \setminus \lbrace 1 \rbrace )$ which includes the points $(0, 1/2$) and $(0,3/2)$ such that the function $f(x,y)=4x^2+xy-\frac{1}{y-1}$ is Lipschitz-continuous

Doubts and concerns: Here $f: \mathbb{R}^2 \to \mathbb{R}$ is a function of two variables. In class we had the following definition of Lipschitz continuity:

Definition: A function $f: I \times U \to \mathbb{R}^m$ is relative to $x$ Lipschitz-continuous $$\iff \exists L \geq 0, \forall t \in I, \forall (x_1,x_2) \in U^2, \text{ s.t. } d(f(t,x_1),f(t,x_2)) \leq L d(x_1,x_2)$$

• So what bothers me is the notion of the relative to (variable). Would you say that $f$ is Lipschitz-continuous if it is Lipschitz-continuous relative to all variables?
• In class we've only dealt with real valued functions $f: \mathbb{R} \to \mathbb{R}$ which are indeed super easy to verify using the criteria of a bound derivative on the given set for example

My approach: After browsing the internet for a while I have found problems such as $f(t,x)$ being solved by partially differentiating with respect to the second variable. Applying this 'self-taught method' I get: $$f(x,y)_y = x - \frac{1}{(y-1)^2} \tag*{(*)}$$ Looking at the given points $(x_1,y_1)=(0,1/2)$ and $(x_2,y_2)=(0,3/2)$ I could say that $x=0$ and $1/2 < y < 3/2 \implies 0< y <1$ which just runs me into more trouble, considering the equation (*)

I would appreciate some basic steps to get me started and maybe some clarification on the definition with the relative to/respect to that bothers me.

-

A function $f \,:\, \mathbb{R}^n \to \mathbb{R}^m$ is lipschitz continuous on $D \subset \mathbb{R}^n$ iff there is an $L \in \mathbb{R}$ such that $$\|f(x) - f(y)\| \leq L\|x-y\| \text{ for all x,y \in D.}$$

Let's assume that $n > 1$, then we can also interpret such an $f$ as a function $$f \,:\, \mathbb{R}\times\mathbb{R}^{n-1} \to \mathbb{R}^m \,:\, (t,(x_1,\ldots,x_{n-1})) \mapsto f((t,x_0,\ldots,x_n)) \text{.}$$ $f$ being lipschitz continuous relative to $t$ then is a weaker concept than (full) lischitz continuity of $f$ as function $\mathbb{R}^n \to \mathbb{R}^m$. It only tells us that for every fixed t, the function $x \mapsto f(t,x)$ is lipschitz continuous, and that one global bound $L$ works for all these functions. But it doesn't give us any bound for $\|f(t_1,x) - f(t_2,y)\|$ for $t_1 \neq t_2$.

However, if $f \,:\, \mathbb{R}^k\times \mathbb{R}^l \to \mathbb{R}^m \,:\, (x,y) \mapsto f(x,y)$ is lipschitz continuous relative to $x$ and to $y$ with lipschitz constants $L_x$ respectively $L_y$, then $f$ is lipschitz-continuous as a function $\mathbb{R}^{k+l} \to \mathbb{R}^m$ because $$\begin{eqnarray} \|f(a_x,a_y) - f(b_x,b_y)\| &=& \|f(a_x,a_y) - f(a_x,b_y) + f(a_x,b_y) - f(b_x,b_y)\| \\ &\leq& \underbrace{\|f(a_x,a_y) - f(a_x,b_y)\|}_{\leq L_x\|a_y - b_y\|} + \underbrace{\|f(a_x,b_y) - f(b_x,b_y)\|}_{\leq L_y\|a_x - b_x\|} \\ &\leq& L'\left( \|a_x - b_x\| + \|a_y - b_y\| \right) \quad L':=\max\{L_x,L_y\}\\ &\leq& L'\left( \|a_x - b_x\| + \|a_y - b_y\| \right) \\ &\leq& L'\left( \|(a_x,0) - (b_x,0)\| + \|(0,a_y) - (0,b_y)\|\right) \\ &\leq& L\sqrt{2} \|(a_x,a_y) - (b_x,b_y)\| \\ &=& L\|(a_x,a_y) - (b_x,b_y)\| \quad L:=\sqrt{2}L' = \sqrt{2}\max\{L_x,L_y\} \text{.} \end{eqnarray}$$ (The last line uses that $a + b \leq \sqrt{2}\sqrt{a^2 + b^2}$. Also, $(a,0)$ is to be read as $a\in\mathbb{R}^k$ with $l$ zeros appended to make it a vector in $\mathbb{R}^{k+l}$, and the same goes for $(0,b)$, only that here $k$ zeros are prepended).

In your case, you're dealing with (full) lipschitz continuity. It's easy to see that the only regions of $\mathbb{R}^2$ which cause trouble are those which either include $y$-values arbitrarily close to $1$, or which are unbounded.

-
Thanks a lot for your answer @fgp, the way you explain this makes it much clearer to me. Since this is my first attempt at solving such a problem, could you please also tell me if my attempt at differentiating $f$ partially with respect to $y$ is a valid step to discuss the Lipschitz-Continuity? (With respect to $y$ in this case I guess) –  Spaced Apr 16 at 19:31
I've added a proof that relative lipschitz continuity "both ways" implies full lipschitz continuity. It follows I think that if the gradient of $f$ exists on some domain $D$, i.e. if all partial derivatives exists on $D$, then those partial derivatives being bounded is a sufficient condition for lipschitz continuity on $D$. –  fgp Apr 16 at 19:56
@Spaced Note, however, that this condition is very far from being necessary! A function doesn't have to be differentiable to be lipschitz-continuous. In fact, the point of lipschitz continuity is mostly to deal with non-differentiable functions - otherwise we could just say "functions with bounded derivative". Here, the simplest way to show lipschitz continuity on some $D$ you pick (which better not include neighbourhoods of $y=0$!), is probably to use the definition directly. –  fgp Apr 16 at 20:00
Thanks a lot for all your effort @fgp, I will accept this as an answer and work with understanding your proof. You're most likely right, applying the definition as you have given it might be the best way to approach this problem. Thanks again for your insight and dedication. –  Spaced Apr 16 at 20:05