Distribution Theory - Uniform Convergence

[MidnightR]

F: C(Omega) -> D'(Omega); F(f) = F_f

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O = Omega

Introduce the notion of convergence on C(Omega) by

f_p -> f as p -> inf in C(O) if f_p(x) -> f(x) for any xEO

Show that then F is a continuous map from C(O) to D'(O)

Hint: Use that if a sequence of continuous functions converges to a continuous junction pointwise then it converges uniformly on any compact.

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Really not sure :S I'd be really grateful if someone could just get me started so I don't sit here staring at it for hours >.<

I've thought about the definition of continuous functions between topological spaces:

A function F: C(O) → D'(0), where C(0) and D'(0) are topological spaces, is continuous if and only if for every open set V ⊆ D'(O), the inverse image

$F^{-1}(V) = \{x \in C(O) \; | \; F(x) \in V \}$

is open.

But not sure how to proceed.

 

[Nino MMP]

Answer:We need to show that for any open set V in D'(O) its inverse image F^{-1}(V) is open in C(O). Let V be an open set in D'(O). We know that V can be expressed as a union of sets {f_i} where each f_i is continuous on Omega. By definition, F^{-1}(V) = \{x \in C(O) \; | \; F(x) \in V \}. Since F(x)=F_f and f_i is continuous on Omega, for any x in F^{-1}(V), we know that F(x) is also continuous on Omega. Since each f_i is continuous on Omega, we also know that each x in F^{-1}(V) is continuous on Omega. Now, since we have a sequence of continuous functions converging to a continuous junction pointwise, we can use the fact that this sequence converges uniformly on any compact. This means that for any x in F^{-1}(V), the sequence of functions converging to F(x) does so uniformly on any compact. Since this holds for all x in F^{-1}(V), we can conclude that F^{-1}(V) is open in C(O) and thus F is a continuous map from C(O) to D'(O).