# Lipschitz Continuity of $\sqrt{1 + \|x\|^2_2}$

Let $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ be a vector-valued function given by $$f(x) = \sqrt{1 + \|x\|^2_2}$$. Show that the gradient of $$f$$ is Lipschitz continuous, in particular with Lipschitz constant $$L = 1$$.

I want to show this directly from $$\|\nabla f(x) - \nabla f(y)\|_2 \leq \|x - y\|_2 \; \forall x, y \in \mathbb{R}^n$$. However, the gradient of $$f$$ is not exactly very clean, and I get lost in the algebra. Is there a simpler way to do this that I'm missing?

• Maybe a direct computation of the gradient and looking for an upper bound... – energy Feb 14 at 2:52

The gradient of your function is $$\nabla f(x)=\frac {2x}{2\sqrt{1+\|x\|^2}}=\frac {x}{\sqrt{1+\|x\|^2}}$$ whose norm is bounded by $$1$$, $$\|\nabla f(x)\|= \frac {\|x\|}{\sqrt{1+\|x\|^2}} \le 1$$ Therefore, your function is Lipschitz continuous with Lipschitz constant equal $$1$$ (just an application of the mean value theorem between any two points in $$\mathbb{R}^n$$).
• @rw435 Does this answer your question? This shows $f$ is Lipschitz, not $\nabla f$ as the question you wrote suggested. Or was it a typo, and Lipschitzness of $f$ was all you wanted? – user744868 Feb 14 at 3:20
• You are right @user744868 I thought it was the Lipschitz continuity of $f$ itself. I will edit my answer – GReyes Feb 14 at 3:27
• Hmmm indeed. To prove $L$-Lipschitz continuity of $\nabla f$, do we not actually need to prove $||\nabla ^2 f|| \leq L$? I overlooked this... – rw435 2 days ago