I'm doing an Introduction to Machine Learning course by myself using some open university coursebook and it has the following question which I've tried to solve, but to no avail:

Let there be a Lipschitz function $g:\mathbb{R}^n\rightarrow \mathbb{R}$.

Meaning the following holds: $|g(x)-g(y)| \leq L \,\cdot \parallel x-y\parallel$ for some constant $L\geq 0$.

Prove that the following function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ defined as:

$f(x) = \parallel\,x\,\parallel ^2 + \, g(x)$ has a global minimum

OK so I want to show that function basically goes to infinity in every direction but I am having trouble proving it. I tried going directly from the definition but I can't figure out what the Lipschitz function gives me (I'm sure I need to use it somehow)

Any help is appreciated! Thank you to anyone who helps


Hint: set $y=0$ and see what the Lipschitz condition does for you.

  • $\begingroup$ hmmm, i tried using the hint, and I got this: (I'm trying to show that the function $f(x)$ is larger than an "always growing" function) : $f(x) \geq \frac{|g(x)-g(0)|^{2}}{L^{2}}+g(x)$ but I can't seem to figure out how to carry on from this. Is there something i'm overlooking? $\endgroup$ – user475680 Mar 3 '15 at 12:35
  • $\begingroup$ You get that $|g(x)-g(0)| \leq L||x||$, so $|g(x)| \leq L||x|| + |g(0)|$. Hence, $f(x) \geq ||x||^2 - L||x|| - |g(0)|$. $\endgroup$ – Tom Mar 3 '15 at 12:55

Potentially there are other methods, but since I am familiar with pde I like to take minimal sequences:

  1. The function is bounded from below since Lipschitz functions only grow linearly. Consequently, $m:=\inf_{x\in\mathbb{R}^n} f(x)$ exists.

  2. Take a sequence $x_k$ such that $f(x_k)\to m$. This sequence exists, due to the first step

  3. Since $f$ is large far out, we may assume that $x_k$ is contained in a ball of radius $R$ around zero.

  4. Thus, a subsequence of $x_k$ (again denoted by $x_k$) converges to some $x$

  5. By continuity, we obtain $f(x)=\lim f(x_k)=m$


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