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Let $f_n$ be a sequence of differentaible functions on $[0,1]$ to $\mathbb{R}$ converging uniformly to a function $f$ on $[0,1]$, Then

  1. $f$ is differentiable and Riemann integrable there

  2. $f$ is uniformly continuous and R-integrable

  3. $f$ is continuous, need not be differentiable on $(0,1)$ and need not be R-integrable on $ [0,1]$,

  4. $f$ need not be continuous.

Well I think 2 is only correct statement as every $f_n$ are differentiable on $[0,1]$ hence uniformly contnous, and as they converge uniformly to $f$ on $[0,1]$ so $f$ is uniformly continuous and every continuous function is rieman integrable..Am I correct?

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  • $\begingroup$ Why do you think that? Do you have a proof of 2.? Do you have counterexamples for the other statements? $\endgroup$ – Ben Millwood Jun 4 '12 at 22:28
  • $\begingroup$ the proof for 2 is as every $f_n$ are differentiable on $[0,1]$ hence uniformly contnous, and as they converge uniformly to $f$ on [0,1] so $f$ is uniformly continous and every continous function is rieman integrable. $\endgroup$ – Marso Jun 4 '12 at 22:30
  • $\begingroup$ Convolve $|x-\frac{1}{2}|$ with smooth approximate identity functions for a counterexample to 1. The uniform limit of Riemann integrable functions is Riemann integrable so 3 is false and 2 is true (uniform limit of continuous is continuous and continuous on compact is uniformly continuous). $\endgroup$ – Chris Janjigian Jun 4 '12 at 22:31
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A simple counterexample to 1 is the sequence $f_n(x)=\sqrt{(x-1/2)^2+1/n}$, which converges uniformly to non-differentiable function $f(x)=|x-1/2|$.

2 is correct: uniform convergence preserves uniform continuity, and uniform continuity implies Riemann integrability. It follows that 3 and 4 are false.

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