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$f_n$ be a sequence of differentiable functions on $[a,b]$ such that $f_n(x)\rightarrow f(x)$ which is riemann integrable consider the statements

  1. $f_n$ converges uniformly.

  2. $f_n^{'}$ converges uniformly.

  3. $\int_{a}^{b}f_n\rightarrow \int_{a}^{b}f$

  4. $f$ is differentiable.

we need to find out which of the statement is not true.

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up vote 1 down vote accepted


For 1., 2., and 4., consider a suitable scaling of the functions $f_n(x)=x^n$ on $[0,1]$.

For 3, think of functions $f_n$ that are $0$ everywhere, except on the interval $(a,a+1/n)$. On this interval, the graph of $f_n$ is a "smooth hump" of area $1$.


You can construct the functions $f_n$ as suggested above for 3., so that for each $n$, there is a $c_n\in(a,a+1/n)$ with $f'(c_n)>1$. Then this sequence will show 2. is false.

For 2., you could also consider the functions $f_n(x)=\sin(nx) /\sqrt n$.

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what are picture should be there in mind because it is not always possible to memorise examples,I have exam knocking at the door :( – La Belle Noiseuse Dec 14 '12 at 14:41
@Kuttus. I'm not sure what you mean. I do think it's better to think of the graphs of the functions here, rather than the equations defining the functions. – David Mitra Dec 14 '12 at 14:59

Hint: The key thing in your problem is the following fact "the sequence $f_n(x)$ converges to a Riemann integrable function $f(x)$ which means $f(x)$ is bounded and has a countable set of discontinuities". That means $f(x)$ does not have to be continuous which implies it does not have to be differentiable.

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Not that it matters for the point you are making, but a Riemann integrable function can also have an uncountable set of discontinuities, as long as the set of discontinuities has Lebesgue measure zero. – user38355 Dec 14 '12 at 15:34
@brom: Thanks for the comment. In fact this is known as Lebesgue Criteria for Riemann integrability, that a function has to be bounded and has dicontinuity on a set of measure zero. But as you know, I do not the background of the OP. – Mhenni Benghorbal Dec 14 '12 at 15:43

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