Drew Brady

I am wondering if the supremum sup trace([S_j (P + L^{-1})S_j^*]^{-1}) converges to sup trace((P + L^{-1})^{-1}) as j \to infinity, where the suprema range over L, which are self-adjoint positive operators with tr(L) <= 1.

Suppose I have a positive semidefinite operator P on l^2 (self-adjoint and bounded). Let S_j denote projection from l^2 onto first j coordinates.

Btw, I agree this isn't serious differential geometry. It did reduce to polar coordinates after all ;)

Anyways, thanks.

Bizarre that in the tangent space this function is convex but not once on the manifold.

(when I said earlier that $1 + y^2$ is convex, I thought you meant as a function from $\Bbb R^n \to \Bbb R$ !!)

Yes, this means that y^2 as map from the circle to R is not convex.

Wait, I'm confused are you saying you agree with my sketch or that you don't agree with what I'd sketched @TedShifrin?

well yes its obvious that as a map [0, \pi] -> R it is not convex

Just to be concrete, you're saying we should essentially consider f(t) = sin^2(t) + 2cos^2(t) for t in [0, pi]

What am I misunderstanding.

I'm confused, 1 +y^2 is convex, though

Yes, that is true...

Parameterizing the function in this space, we get f(theta) = (lambda_1 - lambda_2) cos^2(theta) + constant.

But let me know if something is not convincing in my argument. Basically, we can restrict to the great circle in the plane spanned by the two eigenvectors associated to the two distinct eigenvalues.

Well what I think I showed is that if even two eigenvalues are distinct then you're hosed.

Really? Does positive definiteness matter here? I think distinctness of eigenvalues is all one needs.

Yielding a great circle, and e_i, e_j picked to be assocaited with eigen values which are distinct.

So you can restrict to span{e_i, e_j} intersected with S^{n-1}

yes that's right. (that's what my answer does too)

in which case f looks like a weighted sum of squares.

(my partial response does this)

Absolutely.

Wow. Let me try that again: for every geodesic $\gamma: [0, 1] \to S^{n-1}$, the composite map $f \circ \gamma$ should be convex.

*gamma

What does this mean? It means for a map f: S^{n-1} -> R any geodesic gamam on the n-sphere must yield a convex map f \circ gamma.

Here's the setup. We're on the n-sphere, and considering convex functions on the n-sphere

[More details here, but sadly not receiving much attention: math.stackexchange.com/questions/3290962/…

Ted do you have .a minute to be bothered by a differential geometry question?

Looks like I missed ya tho

but I guess I had a specific question regarding composite maps w/ geodesics

@TedShifrin yeah, they're all paths along great circles essentially right

..yes I know ask, dont ask to ask...but i asked a while ago to crickets, so.

anyone here familiar w/ geodesics on the n-sphere

I'm not super comfortable with Riemannian geometry, so I thought I'd ask here :)

But I wonder if there is a nice way to answer this question on the sphere as I explained above in the geodesic definition.

In case $M = \mathbf{R}^n$, the answer is easy, the answer is that the matrix $A$ needs to be symmetric positive nonnegative definite.

(Indeed if $M = \mathbf{R}^n$ with the usual flat metric, then this reduces to the usual definition.)

Let $M$ be a Riemannian manifold. A map $f : M \to \mathbf{R}$ is called convex (w.r.t. $M$), provided that $f(\gamma(t)) \leq t f(\gamma(0)) + (1 - t) f(\gamma(1))$ for every geodesic $\gamma: [0, 1] \to M$.

Here we consider a different notion of convexity, not the usual one.

Question: for which $n \times n$ matrices $A$ is the map $x \mapsto x^T A x$ convex as a function of $x \in M$, where $M = S^{n-1} \subset \mathbf{R}^n$?

Is there an easy way to see that the Hausdorff 2-dimensional measure of R is 0?

No worries @Ted, I think this works

so the FTC applies by Lebesgue theorem.

because f is Lipschitz hence absolutely continuous

oh, but it does.

And then we conclude that m(R \ A) = 0.

And now the result follows from the fundamental theorem of calculus: m(B) = f(1) - f(0) = \int_0^1 f' = 1.

So supposing that m(B) > 0, we know that this constant must be 1 by Lebesgue density theorem.