Anacardium

In topological vector spaces this is true.

Are convex topological spaces necessarily path connected? I am wondering whether the line joining any two points will be continuous with respect to the underlying topology.

By the same analogy negative convex functions do not exist which is non-constant; although plenty of negative concave function do exist.

Hmm right.

I get your point. Thanks a lot.

So even though it cuts the x-axis, the curve might not.

Because the tangent line is now passing below the curve at any point.

Nice observation. But positive convex functions do exist. Such as $x^2, e^x, e^{-x}$ etc.

Hmm. That is really nice to know. So the question would not be interesting if one replaces convex by concave.

I think the same argument goes through.

You can prove the same thing if the setup of the linked question is changed to positive everywhere.

@leslietownes The question in the link is more general.

So it's a wrong problem as stated.

@leslietownes That's what I came up with. That's why the confusion arose.

This is given as an assignment problem of a course in which I have been assigned as a TA. The problem states "Show that reciprocal of positive convex function is concave."

@leslietownes: I see. But can't we similarly prove reciprocal of a positive convex function is concave?

@leslietownes$:$ Could you please tell me the error that I am making?

So according to the link $-e^{-x}$ is convex. Hence $e^{-x}$ is concave. Is there any mistake here?

It should be concave. Right?

@leslietownes: If $e^x$ is convex what is $-e^x$?

Where did I make mistake?

This tells us that both $e^{x}$ and $e^{-x}$ are simultaneously convex as well as concave which is possible if and only if they are linear functions in $x.$ This is completely absurd.

It may seem a bit silly question to ask but I having hard time getting this. By double derivative test, it is easy to see that both $e^x$ and $e^{-x}$ are convex, as in both the cases the derivatives are strictly increasing function of $x.$ But then I found the following link : math.stackexchange.com/a/532690/512080

@Thorgott: Hmm got it. Thanks.

Consider the set $S$ of all sequences of $0$'s and $1$'s with no consecutive $1$'s. Does there exist any one to one correspondence from $S$ onto the Cantor set?

While expressing points in the Cantor set we usually make use of their ternary representation. As we are removing middle third interval in each step, it follows that all such points have only $0$'s and $2$'s as digits in their ternary representation. But after some trials, it seems to me that binary representation of points in the Cantor set does not have two consecutive $1$'s. Although I cannot able to show prove it. Do anyone have any idea on it?

But the Tietze extension needs the space to be normal.'

You can take U_i's to be disjoint as the underlying space is regular.

Right. You can think of this way as well. Of course details are required. I am not the one who will believe in everything blindly without any reason. It might be obvious to you. But not to me. That's why I am veryfying all the details. There is no harm in doing so.

Your process uses Titeze + Urysohn simultaneously, I guess. Tietze does not specify how the extension should look like except the norm boundedness condition.

Okay. Here the norm is not the sup norm. Rather it is induced by the C_0(G^(0))-valued inner product.

Once you take $\xi_x$ with l^2-norm 1 on $G_x$ Tietze guarantees an extension \xi which is continuous on G and bounded by 1. Now finitely many open set covering the support of \xi_x and let U be the union of those open sets. Then by Urysohn lemma we can define a compactly supported continuous function g on G which takes the value 1 on the support of \xi_x and takes the value 0 outside U. Now I take $\xi_x g.$

My idea is different than that of yours.

What I said that you don't even need bisections to cover the support. What is the use of that.

@DavidGao Not at all. I know that can't happe.

Why?

That's my question. What do I need s to be a homeomorphism on the union?

Bisection means all those subsets E of G such that r and s restrict to homeomorphism.

On the union s might not be a homeomorphism. You just said.

Should I need open bisection at all here?

That's not my question. My question is how to incorporate the finite support of \xi_x inside an open bisection

for obvious reasons.

Any single point has that property.

Which is what you claimed.

So the only part to invesigate is whether any finite set can be included inside a relatively compact open bisection?

I think you can just omit that part. You can find such a continuous function by Urysohn lemma and then multiply it by the Tietze extension of \xi_x.

But on the union s might not be a homeomorphism. Right?

On each of the open bisection s is a homeomorphism.

Now you are covering supp (\xi) by finitely many open bisections.

Easy indeed. Now let me check your argument again.