# Show that $f$ satisfies Lipschitz condition if it is Lipschitz on the lines with rational coordinates

Let $\Omega \subset \mathbb{R}^2$ be nonempty open convex set, $f: \Omega\rightarrow \mathbb{R}$ be a continuous function such that $f_x$ and $f_y$ exist in $A=\{(x,y) \in \Omega : x \mbox{ is rational} \mbox{ or } y \mbox{ is rational}\}$ and are bounded. Show that $f$ satisfies Lipschitz condition.

Since $f$ is convex, for each two points $(x_1,y_1)$, $(x_2,y_2)$ in $A$ thanks to MVT there is some $(\overline x, \overline y)$ in $A$ such that

$$f(x_2,y_2)-f(x_1,y_1)=f_x(\overline x, \overline y)(x_2-x_1)+f_y(\overline x, \overline y)(y_2-y_1)$$

Because $f_x < M_1$ and $f_y<M_2$ in $A$, we have:

$$|f(x_2,y_2)-f(x_1,y_1)|\le |M_1 (x_2-x_1)|+|M_2(y_2-y_1)|$$

From CS inequality we know that

$$|f(x_2,y_2)-f(x_1,y_1)|\le \sqrt{M_1^2+M_2^2} \sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$$

Thus $f$ is Lipschitz continuous on $A$ with constant $M=\sqrt{M_1^2+M_2^2}$.

The only thing left is to show that $f$ is Lipschitz continuous on whole $\Omega$. It seems obvious because of continuity of $f$ and the fact that rationals are dense in reals. But how to prove it formally?

Claim: If $f$ is continuous on a metrix space $X$, and is Lipschitz continuous on a dense subset $A\subset X$, then it is Lipschitz continuous on $X$.

Proof: Given $x,y\in X$, pick two sequences $x_n\to x$, $y_n\to y$ such that $x_n,y_n\in A$. By continuity of $f$ and of the distance function, $$|f(x)-f(y)| = \lim_{n\to\infty} |f(x_n)-f(y_n)| \le L \lim_{n\to\infty} d(x_n,y_n) = L d(x,y)$$ as desired.

Concerning your proof: you should say that MVT is used twice, to go from $(x_1,y_1)$ to $(x_2,y_1)$ and then to $(x_2,y_2)$. Which raises the issue of whether $(x_2,y_1)$ is in the domain: convexity does not guarantee that.

To repair this gap, first work with a square $Q$ contained in $\Omega$. You will obtain that $f$ is $M$-Lipschitz on every such square, with same constant $M$. Then, given two points $p,q\in\Omega$, cover the line segment connecting them by open squares $Q_j$. There is a partition of this segment such that every subsegment belongs to some $Q_j$ (this is a consequence of the Lebesgue number lemma, or can be proved directly). Finally, use the triangle inequality to sum over the partition.

• Thanks. Is the rest of my proof correct? – luka5z May 17 '16 at 16:04
• There is a gap in there: see my expanded answer. – user147263 May 17 '16 at 16:20
• What if I just used the fact that Lipschitz continuity in each variable separately <=> Lipschitz continuity? – luka5z May 17 '16 at 19:29
• You don't have Lipschitz continuity in each variable separately. You only have it on some subset. Presumably, the theorem you are thinking about had assumptions on the domain of the function, which don't include the kind of set described here. – user147263 May 17 '16 at 19:32
• But I was talking about LC on $A$. Clearly it is LC with respect to $x$ on $A$ as well as with respect to $y$. Thus it is LC on $A$. Then we use continuity and extend over whole $\Omega$. Or am I stupid? – luka5z May 17 '16 at 20:27