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Suppose $f_n(x)$ is defined on $[a,b]$, and it converges uniformly to $f(x)$ on $(a,b)$. And the sequences $f_n(a)$ and $f_n(b)$ both converge (say, to points $c$ and $d$ respectively). I want to prove $f_n(x)$ is uniformly convergent on $[a,b]$.

I know that when the goal is to proving convergence, we can combine the points $c,d$ and $f(x)$ to a new function, then the convergence is justified. But I do not know how to proceed for uniform convergence.

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For an arbitrary $\varepsilon > 0$, there are numbers $N_1, N_2, N_3$ so that

  • For all $n \ge N_1$ we have $|f_n(a) - c| < \varepsilon$.
  • For all $n \ge N_2$ we have $|f_n(b) - d| < \varepsilon$.
  • For all $n \ge N_3$ and $x \in (a, b)$ we have $|f_n(x) - f(x)| < \varepsilon$.

Now extend $f(x)$ via $f(a) = c$ and $f(b) = d$ to a function on $[a, b]$. Choosing $N = \max\{N_1, N_2, N_3\}$ now yields $|f_n(x) - f(x)| < \varepsilon$ for all $x \in [a, b]$ and $n \ge N$, i.e. $f_n$ converges uniformly to $f$.

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  • $\begingroup$ Sorry for coming back to this. Since both $N_1$ and $N_2$ depend respectively on $a$ and $b$ (pointwise convergence), saying $N=\max\{N_1,N_2,N_3\}$ makes $N$ totally independent of $x$? I've always wondered if taking max like this avoids this dependency. Most of authors do this but I'd like to confirm. Thanks! $\endgroup$ Commented Feb 12, 2019 at 6:57
  • $\begingroup$ @mate89 That is correct. $\endgroup$ Commented Feb 21, 2019 at 19:27
  • $\begingroup$ @Dominik, hi, if we only know that $f_n$ converges uniformly to $f$ and each $f_n$ is continuous on the open interval, how do we prove that $f_n(a) \to f(a) $? $\endgroup$ Commented Oct 14, 2021 at 0:54

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