How to check the compactness of following sets?

Question

(1) Let $K \subset M_n(\mathbb{R})$ be defined by $$K = \{A \in M_n(\mathbb{R})\mid A = A^T, \ \operatorname{tr}(A) = 1, x^TAx \geq 0 \text{ for all } x \in \mathbb{R}\}$$ Then $K$ is compact.

(2) Let $K \subset C[0, 1]$ (with the usual sup-norm metric) be defined by $$K = \bigg\{f \in C[0, 1]\mid \int_0^1 f(t) \, dt = 1 \text{ and } f(x) \geq 0 \text{ for all } x \in [0, 1]\bigg\}$$ Then $K$ is not compact.

How to prove the above two statements?

For $(1)$ I know that in general the set of positive definite matrices is not closed but what impact does the condition $\operatorname{tr}(A) = 1$ on the compactness?

For $(2)$, how can i utilise Arzela-Ascoli thereom?

Answer 1

One way to look at the first one is to note that the orthogonal matrices are compact, while diagonal matrices with nonnegative diagonal entries summing to $1$ are also compact. This space is more or less the product of those two (specifically, the product of those two and this space are related through a "nice" function). Without the trace requirement, $K$ wouldn't even be bounded.

As for 2, there is a nice easy way to do it, just take a sequence of successively narrower and taller triangular spikes all of which have area 1. Arzela-Ascoli basically just tells you what you can't do, but you don't need to invoke it in the proof.

Answer 2

For the first, you may use Heine-Borel. To see that it is closed, define the following family of functions $$ f_x:\mathbb{R}^{n^2}\to\mathbb{R}, A\mapsto \langle x,Ax\rangle $$ which is continuous, since the inner product is continuous. And define, $$ g:\mathbb{R}^{n^2}\to\mathbb{R}^{n^2}\\ A\mapsto A-A^T $$ A polynomial, and thus certainly continuous.

Then your set is $$ \bigcap_{x\in \mathbb{R}}f_x^{-1}([0,\infty))\cap g^{-1}(\{0\})\cap\text{tr}^{-1}(\{1\}) $$ and thus closed.

For boundedness, note that for symmetric matrices $||A||=\max_{\lambda\in \sigma(A)}|\lambda|$, i.e. the spectral radius. Since the trace is the sum of the eigenvalues, your set is clearly bounded.

Answer 3

Let me try to show you why $K$ in $(1)$ is closed.

Define for $x \in \mathbb{R}^{n}$ the map $T_x: A \rightarrow <x,Ax>$.

Note that:

$1.$ $T$ is Linear and continuous.

$2.$ Set of all positive semidefinite matrices:= $\bigcap_{x \in \mathbb{R}^{n}}T_x^{-1}\{[0, \infty)\}$

$3$. Since each of the sets is closed and it is arbitrary intersection, then it is also closed.

I just have shown for Positive semi-definite.