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Let's say we are in a function space $X\subset \{f:A \subseteq \mathbb{R} \to \mathbb{R}\}$,in which the metric $d(f,g)=\sup_{ A}(|f-g|)$ is well defined($(X,d)$ is a metric space). My doubt is the following: let $\{f_n\}\ $ be a succession of functions in $X$, is the following equivalence correct? $$\{f_n\} \text{ converges uniformly in } A \Leftrightarrow \{f_n\} \text{ converges in } (X,d) $$ I have no problem with $\Leftarrow$ implication. In my opinion $\Rightarrow$ is not true, because for example if $X=C^1([-1,1])$ and $f_n(x)=\sqrt{x^2+\frac1n}$ , than $f_n$ uniformly converges to $|x|$ in $[-1,1]$, but it doesn't converge in the metric space because if it were convergent it would converge to a function $g$ different from $|x|$(because $|x| \not \in C^1([-1,1])$), and by $\Leftarrow$ it would uniformly converge to $g$ in $ A$ and this is absurd because a succession of function cannot uniformly converge to 2 different functions ($g$ and $|x|$) in $A$. I noticed that by Cauchy criterion: $$\{f_n\} \text{ converges uniformly in } A \Leftrightarrow \forall x\in A \forall \varepsilon >0 \exists N_{\varepsilon} \in \mathbb{N}:|f_n(x)-f_m(x)|<\varepsilon \ \ \forall n,m \geq N_{\varepsilon} \Leftrightarrow$$ $$\Leftrightarrow \forall \varepsilon >0 \exists N_{\varepsilon} \in \mathbb{N}:\sup_A(|f_n-f_m|)<\varepsilon \ \ \forall n,m \geq N_{\varepsilon}\Leftrightarrow$$ $$\Leftrightarrow \forall \varepsilon >0 \exists N_{\varepsilon} \in \mathbb{N}:d(f_n,f_m)<\varepsilon \ \ \forall n,m \geq N_{\varepsilon}\Leftrightarrow \{f_n\} \text{ is a Cauchy sequence in } (X,d)$$ So basically the our initial equivalence becomes:

$$\{f_n\} \text{ is a Cauchy sequence in } (X,d) \Leftrightarrow \{f_n\} \text{ converges in } (X,d) $$ Which is true only if $(X,d)$ is complete. So instead of saying that uniform convergence is equivalent to the convergence with respect to the sup metric, it would be more correct to say that it is equivalent to the "Cauchy-ness" with respect to the sup metric. Am I correct?

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The stetement is true if you put it as follows: a sequence $(f_n)_{n\in\Bbb N}$ of elements of $(X,d)$ converges in $(X,d)$ to some function $f\in X$ if and only if it converges uniformly in $A$ to that same function $f$. This holds whether or not $(X,d)$ is complete.

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  • $\begingroup$ Thank you for the answer. So it would be correct to say that a metric space with d as metric, is complete if and only if all the uniformly convergent successions converge to an element of the metric space X. Abusing of terminology, we could say that $X$ is complete only if it's closed with respect to uniform convergence(It remembers a bit the theorem that states that a metric subspace of a complete metric space, needs only to be closed to be complete). $\endgroup$ – Eureka Sep 27 '20 at 16:40
  • $\begingroup$ Yes, that is correct. $\endgroup$ – José Carlos Santos Sep 27 '20 at 16:44

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