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I have a function $f(x)$ from $R^n$ to $R$ and it bounded by a quadratic convex function $g(x)$ for example $ g(x) = || x ||^2_2 $

How do I show that $f(x)$ must have a minimum ( or provide a counter example )? I know that $\lim_{|x| \to \infty} f(x) = \infty$ and that $ f(x) > g(x) $ and intuitively it looks obvious that $f(x)$ has a minimum (unique or not is not important). But how do I show that more formally ?

Edit: based on comments below I'm adding the requirement that $f(x)$ is convex

Edit: Lets assume $ g(x) = || x ||^2_2 $ because we have counter examples without this requirement.

Tomer
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  • One attempt would be this: you first solve the case of $n=1$ which is pretty simple. Then for general $n$ you use the 1-dim case to argue that for each $i=1,..,n$ we must have an $x^{(i)}$ s.t. $\frac{\partial f}{\partial x_i}(x^{(i)})=0$ and which is a local minimum w.r.t. the $x_i$-coordinate. Then you argue that one of those must be a local minimum w.r.t. all coordinates. – StiftungWarentest Jan 29 '23 at 21:55
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    Counterexample: $n=1$, $g(x)=x^2$, $f(x)=\begin{cases}x^2+1&x\ne0\2&x=0\end{cases}$. Then $f$ has no minimum. – Hagen von Eitzen Jan 29 '23 at 22:11
  • I'll modify the question to make f(x) convex as well – Tomer Jan 29 '23 at 22:20
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    It is still possible to fix a counterexample like $g(x,y)=x^2$ and $f(x,y)=x^2+e^y$. What is really important here for the results is that $\lim_{|x| \to \infty} f(x) = \infty$ (sometimes called coercivity) plus the fact that a convex function is continuous on the interior of the domain (the whole $\mathbb{R}^n$ in your case). – A.Γ. Jan 29 '23 at 22:43
  • And if I add a requirement that $ g(x) = ||x|| ^2$ ? – Tomer Jan 30 '23 at 00:50
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    @Tomer The sufficient assumptions to ensure existence of minimum are 1) $f$ is continuous on $\mathbb{R}^n$, and 2) $\lim_{|x| \to \infty} f(x) = \infty$. A proof goes via restricting the space to a large enough compact region and applying Weierstrass theorem on min/max. You may assume instead stronger 1') $f$ is convex on $\mathbb{R}^n$ to guarantee 1), and 2') $f(x)\ge|x|^2$ in replace of 2) – A.Γ. Jan 30 '23 at 08:06
  • @Tomer Related questions here: convex are continuous and coercivity+continuity implies global min. – A.Γ. Jan 30 '23 at 08:20

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