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Composition of Continuous Mapping on Compact Space Preserves Uniform Convergence
Let $(X,d_X)$ be a compact metric space.
Let $(Y,d_Y),(Z,d_Z)$ be metric spaces.
Let $ϕ:Y→Z$ be a continuous mapping.
For all $n\inN$, let $f_n:X→Y$ be a continuous mapping.
Let $f: X \to Y$ be a mapping such that $ ⟨f_n⟩_{n \inN}$ converges to $f$ uniformly on $X$.
Then the sequence $⟨\phi \circ f_n⟩_{n \inN}$ converges to $\phi \circ f$ uniformly on $X$.
composition of uniformly convergence sequence with continuous function, is uniformly convergence?
连续映射的复合保持一致收敛需要$X$是紧致空间.
一个非紧致空间上连续映射的复合不保持一致收敛的例子:
The idea for a counterexample is to have the $f$ be unbounded and $g$ "stretch out" the distance between points near infinity. For example: $f_n:(0,1]\to \mathbb R, f_n (x) = 1/x + 1/n$ and $g(y)=y^2$. Then $f_n$ converges uniformly on $(0,1]$ to $f(x)=1/x$, but $|g(f_n(x))-g(f(x))| = 2/(xn)+1/n^2$ is unbounded for every $n$, so $g \circ f_n$ cannot converge uniformly.
而$|g(f_n(x))-g(f(x))| = 2/(xn)+1/n^2$无界，当$x$趋于0。 |
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