Prove a subset of R^n is compact

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In summary, the function ##f : R^n \setminus \{0\} \mapsto R##, with the condition ##f(\lambda x) = f(x)## for all ##\lambda > 0## and nonzero ##x \in R^n##, attains its global minimum in its domain. This can be proven by considering a compact subset of the domain, such as the norm ball, and using the Weierstrass theorem to show that there exists a global minimum on this subset. Due to the condition on ##f##, this minimum is also the global minimum for the entire domain.
  • #1
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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.
 
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  • #2
The domain of ##f## is neither compact nor closed. Compactness may come into the proof later on, but that would be compactness of a different set, not the domain of ##f##.

Instead, I suggest you focus on the the condition that ##f(x)=f(\lambda x)##. What does that tell us about the how the value of the function varies as we move along a radial line from the origin?
 
  • #3
I think I follow (and understand the point of the hint). Because of the lambda condition, we can create a compact subset (such as the norm ball), which is guaranteed a global minimizer. We argue, since every "ray" from the origin has the same value, we can consider a subset ## \{x : ||x|| \leq 1 \} ##, and use the Weirstrass theorem that there must exist some ##x_0## that is the global minimizer.

Because every "ray" has the same value, the global minimizer on our compact subset is also the global minimizer of the larger, unbounded set. Is that more or less correct?
 
  • #4
That's the general idea, but I don't know what you mean by the norm ball. Are you referring to the ring ##S^1##? That is certainly a compact set. But if you're referring to the set of all points in the domain of ##f## with modulus less than or equal to 1, it is not compact because it does not include the origin.

Also, I suggest avoiding phrases like 'the global minimiser' which implies there is a unique point in the domain where the minimum is obtained. From the ##\lambda## condition we know there are infinitely many such points, comprising at least one ray. In fact the question is posed in a weird way and seems to me to presuppose its conclusion. A better way to state the question is simply 'prove that ##f## has a global minimum', or 'prove that there exists ##x_0## in the domain of ##f## such that ##f(x_0)=\inf A## where ##A## is the image of ##f##'.

Also note that there are several Weierstrass Theorems, covering widely different topics. See here. I'm guessing the one you're referring to is the Bolzano-Weierstrass Theorem.
 
  • #5
Thanks for the feedback, it is very helpful! By norm ball I mean the set ## \{ x \in R^n : ||x||_{2} < c \} ## where c is some constant. It's the term our professor has been using, though the notation involved has been a bit scattered. This is largely an applied convex optimization course, with the first few weeks being a mad dash through the theory, and I suspect the normal formalities of analysis are being thrown to the wind. I suspect the odd phrasing is due to the questions being pulled from a variety of texts, and he tried to "normalize" the wording slightly, but I couldn't really say for sure.
 

Related to Prove a subset of R^n is compact

1. What is a subset of R^n?

A subset of R^n is a set of points in n-dimensional Euclidean space. It can be thought of as a smaller or more specific version of the entire n-dimensional space.

2. What does it mean for a subset of R^n to be compact?

A compact subset of R^n is one that is closed and bounded. This means that the subset contains all of its boundary points and is limited in size, with a finite distance between its farthest points.

3. How can I prove that a subset of R^n is compact?

To prove that a subset of R^n is compact, you can use one of several methods, such as the Heine-Borel Theorem, the Bolzano-Weierstrass Theorem, or the sequential compactness property. These methods involve showing that the subset is closed and bounded, or that every sequence in the subset has a convergent subsequence.

4. Why is it important to prove that a subset of R^n is compact?

Proving that a subset of R^n is compact is important because it allows us to make certain conclusions about the behavior of the subset. For example, we can use compactness to guarantee that a continuous function on the subset has a maximum and minimum value, or that a sequence in the subset must converge to a point in the subset.

5. Can a subset of R^n be both open and compact?

No, a subset of R^n cannot be both open and compact. In fact, a subset of R^n is compact if and only if it is closed and bounded. This means that a compact subset cannot have any holes or gaps in it, and it must have a finite size.

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