Let $f_n, f:[a, b] \rightarrow \mathbb{R} $ for $n=1,2,3,...$ and suppose $\lim_{n\rightarrow \infty} f_n(t) =f(t) $, ie for all $t \in [a,b], \epsilon>0$ there exists $N(t, \epsilon) \in \mathbb{N} $ such that for $n>N, |f_n(t) - f(t) |<\epsilon. $

Consider the set $A = \lbrace N(t, \epsilon) :t \in [a, b] \rbrace$. Then for all epsilon greater than zero why wouldn't it be true for $ n>\sup A$ (which is an integer), that $|f_n(t) - f(t) |< \epsilon$, for all $t\in[a, b] $? (ie the sequence is uniformly convergent)

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    $\begingroup$ What if $\sup A = \infty$? $\endgroup$ – Sangchul Lee Jun 13 '17 at 7:08
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    $\begingroup$ how could that be? By pointwise convergence there is an $N_t$ for each $t$ in the interval so wouldnt the sup (max?) be one of these $N_t's$? $\endgroup$ – Rlanls Jun 13 '17 at 7:13
  • $\begingroup$ @Arthur, No. Improving pointwise convergence to uniform convergence occurs only in rare occasions. For instance, Dini's theorem tells that if increasing sequence of continuous functions on a compact set converges pointwise to a continuous function, then the convergence is indeed uniform. $\endgroup$ – Sangchul Lee Jun 13 '17 at 7:14
  • $\begingroup$ You are taking supremum of $N(t, \epsilon)$ over uncountably many $t$'s. There is no reason to expect that this is finite. For instance, if $f_n, f : [0, 1] \to \mathbb{R}$ are such that $f_n(t) = t^n$ and $f$ is the pointwise limit, then for $t \in (0, 1)$ and $\epsilon \in (0, 1)$ we have the lower bound $$N(t, \epsilon) \geq \left\lceil \frac{\log\epsilon}{\log t} \right\rceil - 1$$ and hence $\sup A = \infty$. $\endgroup$ – Sangchul Lee Jun 13 '17 at 7:16
  • $\begingroup$ ok so the uncountability is the issue. what assumptions would allows the uncountable set of $N(t,\epsilon)$s to have a supremum? $\endgroup$ – Rlanls Jun 13 '17 at 11:43

Consider the sequence $f_n(t)=t^n$ on $[0,1]$ then $f(t)=0$ for $t\in [0,1)$ and $f(1)=1$. Then $$\sup_{t\in[0,1]}|f_n(t)-f(t)|=\sup_{t\in[0,1)}t^n=1$$ and the convergence is not uniform in $[0,1]$.

What is $N(t,\epsilon)$ in this case? What is $\sup \{ N(t, \epsilon) :t \in [a, b] \}$?


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