Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be given by $f(x)=\sqrt{1+||x||^2}$. Show that $f\in C_1^{1,1}$

Question

Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be given by $f(x)=\sqrt{1+||x||^2}$. Show that the $f\in C_1^{1,1}$.

Relevant definitions and theorems:

$[C_L^{1,1}]$ is the class of functions with Lipschitz gradient with constant $L$.

A function is Lipschitz continuous if $$||\nabla f(x)-\nabla f(y)||\leq L||x-y||$$ for any $x,y\in\mathbb{R}^n$.

Theorem: Let $f$ be a twice continuously differentiable function over $\mathbb{R}^n$. Then the following two claims are equivalent:

1. $f\in C_L^{1,1}(\mathbb{R}^n)$
2. $\|\nabla^2f(x)\|\leq L$ for any $x\in\mathbb{R}^n$.

My attempt:

I am wondering if this problem is analogous to the function $f:\mathbb{R}\rightarrow\mathbb{R}$ where $f(x)=\sqrt{1+x^2}$. Calculating the second derivative of this function, I obtain $$f''(x)=\frac{1}{(1+x^2)^{3/2}}\leq1\quad\forall x$$ which by the above theorem means that $f\in C_1^{1,1}$.

However, I am not sure how to differentiate the function involving the norm of $x$.

Could I get some pointers in the right direction for this problem? Thank you.

Answer 1

Let $g(y) = \sqrt{1+y}$ and let $h(x) = \|x\|^2$. Then $f=g \circ h$, and thus

$$\nabla f(x) = g'(h(x)) \nabla h(x) = \frac{1}{2\sqrt{1+\|x\|^2}} \ 2x = \frac{x}{\sqrt{1+\|x\|^2}}$$ and \begin{align} \nabla^2 f(x) &= \left[\frac{\partial}{\partial x_i} (\nabla g(x))_j\right]_{ij} \\ &= \left[\frac{\partial}{\partial x_i} \frac{x_j}{\sqrt{1+\|x\|^2}}\right]_{ij} \\ &= \left[\delta_{ij} \frac{1}{\sqrt{1+\|x\|^2}} - \frac{x_i x_j}{(1+\|x\|^2)^{3/2}}\right]_{ij} \\ %&= \left[\frac{\delta_{ij}(1+\|x\|^2) - x_i x_j}{(1+\|x\|^2)^{3/2}}\right]_{ij} %\\ &= \frac{1}{\sqrt{1+\|x\|^2}} I - \frac{1}{(1+\|x\|^2)^{3/2}} xx^\top \end{align}

(Note that when $n=1$, we have $f''(x) = \frac{1}{\sqrt{1+x^2}} - \frac{x^2}{(1+x^2)^{3/2}} = \frac{1}{(1+x^2)^{3/2}}$ as you have already computed.)

The eigenvalues of $\nabla^2 f$ are $\frac{1}{\sqrt{1+\|x\|^2}}$ (with multiplicity $n-1$, corresponding to vectors orthogonal to $x$) and $\frac{1}{\sqrt{1+\|x\|^2}} - \frac{\|x\|^2}{(1+\|x\|^2)^{3/2}} = \frac{1}{(1+\|x\|^2)^{3/2}}$ (with multiplicity $1$, corresponding to eigenvector $x$).