4
$\begingroup$

In exercise 1 of Charles Chapman Pugh's Real Mathematical Analysis, is a question of when pointwise and uniform convergence are 'equivalent' for metric spaces $M$ and $N$ and a sequence of functions $f_n: M \to N.$ Under my understanding, this would mean that any sequence of functions $f_n \to f$ would immediately imply $f_n \rightrightarrows f$ for a particular combination of $M$ and $N$, since the converse is already true.

What I have found:

If $M$ is finite, then the convergences are equivalent. In other words, $$\forall \epsilon>0,\forall x \in M, \exists L(\epsilon,x) \in \mathbb{N} \hspace{3mm} \text{s.t.} \hspace{3mm} d_M(f_n(x), f(x)) < \epsilon \hspace{3mm} \text{if} \hspace{3mm} n \ge L(\epsilon,x)$$ implies uniform convergence since $L$ and fixed $\epsilon > 0$ is a function on a finite space so $\max_{x}{L(\epsilon,x)} = L(\epsilon)$ exists. Using this maximum, we swap the $x$ and $L$ quantifiers giving us the definition for uniform convergence for $f_n$.

In trying to find more examples or a totality of every combination is where I am stuck. One lead is the fact that if $M$ is compact and $f_n$ is pointwise equicontinuous and pointwise bounded, then $f_n$ being pointwise convergent would imply uniform convergence by generalized Arezelà-Ascoli theorem. Yet, I can't think of any properties to attach to a metric space which would force every sequence of pointwise convergent functions to be pointwise equicontinuous and pointwise bounded.

$\endgroup$
  • 1
    $\begingroup$ You might be interested in Dini's theorem. $\endgroup$ – Surb Jun 27 '17 at 16:28
  • $\begingroup$ While we are on the subject of Italy and to be particular, I will further note that pointwise convergence and uniform convergence are equivalent on a compact set $K$ whenever the sequence of functions forms an equicontinuous and pointwise bounded family there as well. $\endgroup$ – Matt A Pelto Jul 2 '17 at 0:47
2
$\begingroup$

If $M$ is finite they are equivalent.

If $M$ is not finite, they are not equivalent, unless $N$ is empty or contains one element. To see why they are not equivalent in the nontrivial case, take points $a,b \in N$, and points $x_1,x_2,x_3,\dots \in M$. Define $f$ by $f(x) = a$ for all $x \in M$. Define $f_n$ by $f_n(x) = a$ for all $x \in M\setminus\{x_{n+1},x_{n+2},x_{n+3},\dots\}$ and $f_n(x_{n+j}) = b$ for $j \ge 1$. Then $f_n \to f$ but not uniformly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.