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If a sequence {$f_n$} pointwise convergent on a finite set $D$ which is a subset of $\mathbb R$.Then can we say that the convergence is uniform?

If the answer is yes then I have some problem here:

Let us suppose that $D$={$0$,$1$} and $f_n(x)$=$x^n$ then $\lim_{n\rightarrow \infty} f_n(x)=f(x)$

where $f(x) = \begin{cases} 0 & \text{ if } x=0\\ 1 & \text{ if }x=1 \\ \end{cases}$

so $f_n(x)$ pointwise convergent on$D$.But if $M_n= \max(\{|f_n(x) - f(x)| \ \big| \ \ x \in D\})$ then $\lim_{n\rightarrow \infty} M_n=1$ from which I can say that $f_n(x)$ is not uniformly convergent on $D$.Am I right or wrong?somebody please help me.

liesel
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    Note that in the case where $D={0,1}$ and $f_n(x)=x^n$ we have that $f_n(x)=x$ for all $x$, therefore $f(x)=x$ as well. If that's not uniform convergence, I don't know what is! – Asaf Karagila Apr 22 '14 at 11:32
  • Pointwise convergence does not imply uniform convergence. Not even if you have pointwise convergence everywhere (like in your example). – Thomas Apr 22 '14 at 11:36
  • @TedShifrin Ah yes, my bad. On second thought it is easy to come up with a counterexample when the set is compact, e.g. make $f_n$ equal to $1$ on a smaller and smaller set, but $0$ elsewhere. However, it is certainly true when the set is finite, as in this question. – Caleb Stanford Apr 22 '14 at 13:16

1 Answers1

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Yes, pointwise convergence on a finite set is uniform, because if $D = \{x_1, x_2, \ldots, x_k\}$, then for any $\epsilon > 0, \exists N_1, N_2, \ldots N_k \in \mathbb{N}$ such that $$ |f_n(x_i) - f(x_i)| < \epsilon \quad\forall n\geq N_i $$ So just take $N = \max\{N_1, N_2, \ldots, N_k\}$ and you see that $$ \sup_{x\in D} |f_n(x) - f(x)| < \epsilon \quad\forall n\geq N $$ In your example, as pointed out in the comments, $$ f_n(x) = f(x) \quad\forall x\in D $$ so $M_n = 0$ for all $n$. So, $M_n$ does not converge to $1$.

Prahlad Vaidyanathan
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