#function space

[cauchyfeelings]

Feigenbaum 😭😭😭

[tilde-he]

Eq(f,·) is upper hemicontinuous under some reasonable conditions

The equalizer of f and g is defined as: Where f,g : X→Y Eq(f,g) = {x∈X|f(x)=g(x)}

Let X be a locally compact Hausendorff, and Y some topological space. (reminder/explanation because I had to look these up myself: Hausendorff means for any 2 points in the space, there exists a neighborhood of each of the two points such that the intersection of the neighborhoods is empty. This is true for lots of spaces one would probably look at. Tbh I don’t exactly know what locally compact means, but I know Hausendorff and compact implies locally compact, and that compact means any open cover has a finite subcover. ) Let F be the space of continuous functions from X to Y, with the Compact-open topology. Let F×X have the product topology. Let f be some function in F (and therefore be continuous). Let f be ubiquitous, i.e. for any g in F, Eq(f,g) is not empty.

Define h:F→Powerset(X) as h=Eq(f,·). i.e. h(g)=Eq(f,g).

Claim: h is upper hemicontinuous.

Consider g_n a convergent sequence in F which converges to g_inf, and a convergent sequence x_n such that for each n, x_n ∈ g_n , and the x_n converge to x_inf .

For all n, x_n in h(g_n) , i.e. x_n in Eq(f,g_n) , i.e. f(x_n)=g_n(x_n) . So for all n, g_n(x_n)=f(x_n).

Because the g_n converge to g_inf in F , and the x_n converge to x_inf in X, the sequence (g_n,x_n) converges to (g_inf,x_inf) in F×X. It is a theorem that eval:C(X,Y)×X→Y with eval(g,x)=g(x) is continuous when C(X,Y) has the compact-open topology and X is locally compact Hausendorff, and C(X,Y)×X has the product topology. So, eval:F×X→Y is continuous. So, because (g_n,x_n) converges to (g_inf,x_inf), eval(g_n,x_n) converges to eval(g_inf,x_inf). eval(g_inf,x_inf)=g_inf(x_inf), and eval(g_n,x_n)=g_n(x_n).

From before, for all n, g_n(x_n)=f(x_n).

f(x_n) converges to f(x_inf) because x_n converges to x_inf, and f is continuous.

So, g_n(x_n) converges to g_inf(x_inf), but g_n(x_n)=f(x_n), and f(x_n) converges to f(x_inf). Therefore, g_inf(x_inf)=f(x_inf). So, x_inf is in Eq(f,g_inf). So, x_inf is in h(g_inf).

So, we have shown that for any g_n converging to g_inf in F, and any x_n in h(g_n) , with x_n converging to x_inf in X, x_inf is in h(g_n).

Therefore, h is upper hemicontinuous (as this is the definition of upper hemicontinuity).

But wait, none of this required that f be ubiquitous, so that isn’t actually required. f can be any function in F.

But I learned all this in order to try to understand things about ubiquitous functions, so I put it there. Also, it is nice that it being ubiquitous means that the output of h is never the empty set.

So, yeah, this proof, and this post, are done.

Oh, also, h(g) is always closed. If X is compact, then h(g) is compact.

Please ask if you have any questions about this proof (or if my notation is unclear). Also, if you read this whole thing, and no one else seems to have done so already, I’d appreciate it if you let me know that someone read this whole thing. Thanks.

[card-outdoor-artificial-e]

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