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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.

3 votes
0 answers
227 views

$f(x)>0$ and $f(y)>0$ implies $f(x+y)>0$, then there must exist an linear function $g$ such ...

Background: Let $x,y\in\mathbb (0,+\infty)^n$. $f$ is a continuous function on $\mathbb R^n_+=(0,+\infty)^n$. Consider the following condition (1), the sign of $f(x+y)$ is dependent on the sign of $f …
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4 votes
1 answer
1k views

Is the space of continuous functions from Polish space to Polish space Polish?

Theorem 4.19 in Kechris' Classical Descriptive Set Theory says that the space of continuous functions from a compact metric space to a Polish space is Polish. It is therefore obvious that the space of …
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  • 263
-1 votes
2 answers
390 views

$X$ is Polish and $N$ is countable. Is $N^X$ Polish? [closed]

$X$ is a separable, completely metrizable topological space equipped with its sigma algebra of Borel sets. $N$ is a countable space. $X^N$ is the collection of all mappings from $N$ to $X$. It is equi …
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  • 263
1 vote
0 answers
385 views

The perturbation of a convex function can also be convex?

$ W^{1,\infty}(D)\ni f:D\to\mathbb R, (x,y)\mapsto f(x,y)$, is a strictly increasing on both dimensions (i.e. if $x_1>x_2$ then $f(x_1,y)>f(x_2,y)$), lipschitz continuous function defined on a convex …
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  • 263