- #1

Zerox5f3759df

## Homework Statement

Let the function ## f : R^n \setminus \{0\} \mapsto R ## be continuous satisfying ## f(\lambda x) = f(x) ## for all ## \lambda > 0 ## and nonzero ## x \in R^n ##. Prove f attains its global minimizer in its domain.

We are given a hint that the Weierstrass theorem states that a continuous functions attains its global minimizer over compacts sets of ##R^n##.

## Homework Equations

The book we are working out of doesn't give definitions for most things in real analysis, but the definitions I'm familiar with that seem relevant are the following:

A set is closed iff its complement is open.

A set is closed if it contains its limit points.

A set is bounded if there exists ##M > 0## such that ##|a| \leq M## for all ##a \in A##.

A set K is compact if every sequence in K has a sub sequence that converges to a limit that is also in K.

## The Attempt at a Solution

I'm struggling with the general strategy to employ. I've tried showing that the set is closed and bounded by arguing that the set contains its limit points (especially around f(0)), and that the condition ## f(\lambda x) = f(x) ## gives us a bound of ## f(\lambda x)## because no matter the x we select, f(x) is bounded by ## f(\lambda x)##.

I think one part of my confusion is that I'm having trouble mentally relating how the lambda constraint on f(x) impacts the set of x that I am working on. Additionally, I'm not convinced that arguing that f(0) is a limit point gives me that the set is closed. I'm not sure if I'm overlooking something simple and making this more complicated than it needs to be.

Overall I'm hoping for suggestions and strategies for these types of proofs, since I need to prove similar claims for matrix forms of quadratics after this, as well as seeing if I'm missing a fundamental property or theorem that is needed.