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In GF Simmon's book on Topology, I came across the definitions of point wise convergence of a sequence of functions and uniform convergence of a sequence of functions.

For all future references, $C[X,R]$ is the set of continuous bounded real functions defined on the metric space $X,$ and $B$ is the subset of all bounded real functions on $X.$

Let $\{f_n\}$ be a sequence of real functions defined on $X.$

Pointwise convergence is when if for each point $x\in X,$ the sequence of real functions $\{f_n(x)\}$ converges to a function $f(x)$, i.e., at each $x,$ for a given $\varepsilon$, there exists a natural $N$ such that for all $n>N, $ $|f_n(x)-f(x)|<\varepsilon$.

Uniform Convergence is when the $\varepsilon$ doesn't depend on $x.$ Which means that a natural $N$ can be found for each $\varepsilon$ such that it works for all $x\in X.$

In terms of convergence of the sequence of functions to a function, Pointwise means that for a given $\varepsilon$ you will find an $N$ for each $x\in X.$ Uniform means that for a given $\varepsilon$ you will find an $N$ which works for all $x\in X.$ Am I right?

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  • $\begingroup$ Writing $f_n(x)$-$f(x)$ rather than $f_n(x)-f(x)$ is not proper MathJax usage. See my edits to this question. $\endgroup$ – Michael Hardy Nov 21 at 18:14
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    $\begingroup$ Dear @MichaelHardy, Thanks for your kindness. I am learning. $\endgroup$ – Krishan Nov 22 at 6:14
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Well, strictly speaking, the quantifiers in your last paragraph are not okay. Actually, it should be: "Pointwise mean that for a $\varepsilon>0$, and for each $x\in X$, there exists an $N\in\mathbb N$ such that...". Can you see the difference? We talk first about the point $x\in X$. Your understanding of uniform convergence is correct.

With respect to the title of you question, these kind of convergence are not the same, even when $X$ is compact. Consider $X=[0,1]$, the sequence $f_n(x)=x^n$ converge pointwise to a function which is not even continuous, despite of that, the sequence does not have a uniform limit (since the uniform limit of continuous functions is continuous).

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    $\begingroup$ There is no difference between "$\forall \varepsilon > 0$, $\forall x \in X$, ..." and "$\forall x \in X$, $\forall \varepsilon > 0$..." The quantified definition of pointwise convergence in the OP's post is correct. $\endgroup$ – TheSilverDoe Nov 21 at 18:44
  • $\begingroup$ He wrote: "you will find an $N$ for each $x\in X$". That can be written as $\exists N\forall x$, which is not the same that $\forall x\exists N$. $\endgroup$ – Emmanuel C. Nov 21 at 18:57
  • $\begingroup$ I did not talk about the "for all" quantifiers, I referred to the "existence" one, before the "each x\in X". $\endgroup$ – Emmanuel C. Nov 21 at 18:59
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    $\begingroup$ "You will find an $N$ for each $x \in X$". is not $\exists N, \forall x$. That would be "you will find an $N$ such that for each $x \in X$"... $\endgroup$ – TheSilverDoe Nov 21 at 19:37
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    $\begingroup$ Dear @EmmanuelC, I believe what you have written in your first comment is just right. I had meant just that: there exists an N for each x in X, and not for all x in X there exists an N. $\endgroup$ – Krishan Nov 22 at 6:28

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