# Does Lipschitz continuity imply L-smooth?

A function is Lipschitz with constant $$L$$ if $$\forall x,y \in \mathbb{R}^d$$ it satisfies

$$$$\|f(x)-f(y)\| \leq L\|x-y\|$$$$

A function is $$L$$-smooth if $$\forall x,y \in \mathbb{R}^d$$ it satisfies

$$$$\|\nabla f(x) - \nabla f(y)\| \leq L\|x-y\|$$$$

My question is, does the first equation automatically imply the second?

I know that if a function is Lipschitz, then its gradients are bounded everywhere. To see this, rearrange the first equation:

$$$$\frac{\|f(x) - f(y)\|}{\|x-y\|} \leq L$$$$

Then define $$h=x-y$$ and take the limit,

$$$$\lim_{h \to 0} \frac{\|f(y+h) - f(y)\|}{\|h\|} = \|\nabla f(y)\| \leq L$$$$

This implies the following bound on $$f(y)$$:

$$$$f(y) \leq f(x) + L\|y-x\|$$$$

I also know that $$L$$-smooth implies the following bound:

$$$$f(y) \leq f(x) + \nabla f(x)^\top (y-x) + \frac{L}{2}\|y-x\|^2$$$$

These last two equations look similar. In particular, $$\frac{L}{2}\|y-x\|^2 \geq 0$$, so I can add it to the first bound and get a true statement:

$$$$f(y) \leq f(x) + L \|y-x\| + \frac{L}{2}\|y-x\|^2$$$$

And now this looks very similar to the second bound. Am I right to think there's a connection here? Does Lipschitz continuity always imply $$L$$-smooth?

• Lipschitz continuity does not even imply existence of derivative. Example, $f(x)=|x|$. Commented Feb 6, 2020 at 5:21

If $$d = 1$$ then a weaker version of your question is whether $$f$$ Lipschitz implies $$f'$$ Lipschitz. For a counterexample, take $$f(x) = \int_0^x g(t)dt$$ for any bounded function $$g$$ with unbounded derivative. Then $$f' = g$$ is bounded, so $$f$$ is Lipschitz, but $$f'$$ is not Lipschitz since $$f'' = g'$$ is unbounded. In fact the converse is not true either: $$f'$$ Lipschitz need not imply $$f$$ Lipschitz. For instance, take $$f(x) = x^2$$.