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Suppose

  1. $X$ is compact
  2. $f_n:X\to\mathbb{R}$ is continuous for all $n\in\mathbb{N}$
  3. $f_n\to f$ pointwise, and $f$ is also continuous
  4. the sequence $\left\{f_n\right\}$ is uniformly bounded on $X$

Does it follow that the convergence is uniform?

(Note the first three conditions are the same as the hypotheses of Dini's theorem, minus monotonicity.)

I've tried the standard counterexamples showing the conjunction (1)+(2)+(3) is not sufficient for uniform convergence, but those examples aren't uniformly bounded.

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Let $f_n(x)=nxe^{-nx}$ on $[0,1]$. Then $f_n\to 0$ pointwise. Note that $f_n'(x)=ne^{-nx}-n^2x^2e^{-nx}$ which is $0$ at $x=\frac1n$. Note that $f_n'$ is positive below and negative above $\frac1n$, so $f_n$ is maximized at $\frac1n$. Note that $f_n(\frac{1}{n})=\frac{1}{e}$.

So no uniform boundedness is not enough.

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  • $\begingroup$ Thanks. I was thinking about the counterexamples that work as well to show one can't exchange limit and integral: those always blow up, in addition to sliding to the boundary. I should've realized it's just the sliding that matters for my question. $\endgroup$ – symplectomorphic Aug 12 '16 at 7:05

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