Prove that a Lipschitz mapping is continuous

Question

A function $f:(M,\ d)\to(N,\ \rho)$ ρ) is called Lipschitz if there is a constant $K<\infty$ such that $\rho(f(x), f(y))\le Kd(x, y)$ for all $x,\ y\in M$. Prove that a Lipschitz mapping is continuous.

My attempt:

To show that $f(x)$ is continuous, it suffices to show that the inverse image of every basis element $B_\rho(f(x),\ \epsilon)$ (I am assuming that $f$ is onto; is this assumption required?) where $x\in M$ is open, i.e., $f^{-1}(B_\rho(f(x),\ \epsilon))$ is open.

Let $x_0\in f^{-1}(B_\rho(f(x),\ \epsilon))$

$\implies f(x_0)\in B_\rho(f(x),\ \epsilon)$

$\implies\rho(f(x),\ f(x_0))<\epsilon$ ---------- (1)

We need to find a $\delta>0$ s.t. $B_d(x_0,\ \delta)\subset f^{-1}(B_\rho(f(x),\ \epsilon))$.

Let $y\in B_d(x_0,\ \delta)$

$\implies d(x_0,\ y)<\delta$

$\implies Kd(x_0,\ y)<K\delta$

So, the given Lipschitz condition implies $\rho(f(x_0),\ f(y))<K\delta$ ----------- (2)

We need to prove that $\rho(f(x),\ f(y))<\epsilon$.

By the triangle inequality, $\rho(f(x),\ f(y))\le\rho(f(x),\ f(x_0))+\rho(f(x_0),\ f(y))<\epsilon+K\delta$.

Thus the RHS of my inequality is too big for any positive value of $\delta$. Where have I gone wrong?

Answer 1

Let $f : (M, d) \rightarrow (N, \rho)$ be a $K-$Lipschitz mapping.

Let $\epsilon > 0$, and $\eta = \epsilon/K$. Let $x \in M$.

For all $y \in M$ such that $d(x,y) \leq \eta$, you have $\rho(f(x), f(y)) \leq K \times (\epsilon/K) = \epsilon$.

Doesn't that prove that $f$ is continuous in $x$ ?

Answer 2

The idea is you want to keep away from the boundary of your initial open ball. So, take whichever is smaller, the distance to your point $f(x)$ or your distance to the boundary of your open ball, and halve it.

Let $\delta = \min\left\{ \dfrac{\rho(f(x),f(x_0))}{2K}, \dfrac{\epsilon-\rho(f(x),f(x_0))}{2K} \right\}$

Note:

$$\delta = \dfrac{\rho(f(x),f(x_0))}{2K}$$ when $$\rho(f(x),f(x_0)) < \dfrac{\epsilon}{2}$$

And $$\delta = \dfrac{\epsilon - \rho(f(x),f(x_0))}{2K}$$ when $$\rho(f(x),f(x_0)) > \dfrac{\epsilon}{2}$$

So, you get $$\rho(f(x),f(y)) \le \rho(f(x),f(x_0))+\rho(f(x_0),f(y)) < \rho(f(x),f(x_0))+K\delta$$

There are two cases. When $\rho(f(x),f(x_0)) < \dfrac{\epsilon}{2}$, you have:

$$\rho(f(x),f(x_0))+K\delta = \dfrac{3\rho(f(x),f(x_0))}{2} < \dfrac{3}{2}\cdot \dfrac{\epsilon}{2} = \dfrac{3}{4}\epsilon < \epsilon$$

When $\rho(f(x),f(x_0))\ge \dfrac{\epsilon}{2}$, you have:

$$\rho(f(x),f(x_0))+K\delta = \dfrac{\epsilon+\rho(f(x),f(x_0))}{2} < \dfrac{\epsilon + \epsilon}{2} = \epsilon$$

Answer 3

I solved it myself!

To show that $f(x)$ is continuous, it suffices to show that the inverse image of every basis element $B_\rho(f(x),\ \epsilon)$ where $x\in M$ is open, i.e., $f^{-1}(B_\rho(f(x),\ \epsilon))$ is open.

Let $x_0\in f^{-1}(B_\rho(f(x),\ \epsilon))$

$\implies f(x_0)\in B_\rho(f(x),\ \epsilon)$

$\implies\exists\epsilon'>0$ s.t. $B_\rho(f(x_0),\ \epsilon')\subset B_\rho(f(x),\ \epsilon)$ ---------------- (1)

We claim that $\displaystyle B_d\left(x_0,\ \frac{\epsilon'}{K}\right)\subset B_\rho(f(x),\ \epsilon)$.

Let $\displaystyle y\in B_d\left(x_0,\ \frac{\epsilon'}{K}\right)$

$\displaystyle\implies d(x_0,\ y)<\frac{\epsilon'}{K}$

$\implies Kd(x_0,\ y)<\epsilon'$

$\implies\rho(f(x_0),\ f(y))<\epsilon'$

$\implies f(y)\in B_\rho(f(x_0),\ \epsilon')$

$\implies f(y)\in B_\rho(f(x),\ \epsilon)$ [Using (1)]

$\implies y\in B_\rho(f(x),\ \epsilon)$

Thus, $\displaystyle B_d\left(x_0,\ \frac{\epsilon'}{K}\right)\subset B_\rho(f(x),\ \epsilon)$.

In addition, $\displaystyle x_0\in B_d\left(x_0,\ \frac{\epsilon'}{K}\right)$.

So, there is a basis element $\displaystyle B_d\left(x_0, \frac{\epsilon'}{K}\right)$ s.t. $\displaystyle x_0\in B_d\left(x_0, \frac{\epsilon'}{K}\right)\subset f^{-1}(B_\rho(f(x),\ \epsilon))$

So, $f^{-1}(B_\rho(f(x),\ \epsilon))$ is open in $M$.

So, $f$ is continuous.

QED

Answer 4

For any $x\in M$, consider the neighborhood $B_\delta^d(x)$ with $\delta < \frac{\epsilon}{2K}$. Because metrics must satisfy the triangle inequality

$$\forall z_1, z_2 \in B_\delta^d(x):\, d(z_1, z_2) \leq d(z_1, x) + d(z_2, x) < \epsilon/K$$ and because $f$ is Lipschitz, $$\rho(f(z_1), f(z_2)) \leq Kd(z_1, z_2) < \epsilon$$ meaning that $f(B_\delta^d(x)) \subset B_\epsilon^\rho(f(x))$, completing the proof.