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I'm asked to give an example of a sequence $\left\{f_{n}\right\}$ of functions and a function $f$ such that

(a) $f_{\mathrm{n}} \in \mathscr{R}[a, b]$ for every positive integer $n$

(b) $f \in \mathscr{R}[a, b]$

(c) $\lim _{n \rightarrow \infty} \int_{a}^{b} f_{n}=\int_{a}^{b} f$

(d) $\lim _{n \rightarrow \infty} f_{n}(x)$ does not exist for any $x \in[\mathrm{a}, \mathrm{b}]$

I think this is equivalent to giving a sequence of function $\{g_n\}$ such that $g_n$ converges to $g$ uniformly but $g_n'$ doesn't exist. But I can't find such an example

9pound15pence
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    It's not equivalent as you claim, because if you ask that $g_n \to g$ uniformly then you've violated part (d). – Ethan Dlugie Apr 19 '20 at 03:48
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    Unless I'm missing something, $f$ is irrelevant here. Once you've found $f_n$ such that $\lim_n \int f_n$ exists, then you can just take some adequate constant as $f$. – Reveillark Apr 19 '20 at 03:50
  • @EthanDlugie Here I use $g_n$ to represent the antiderivative of $f_n$, in other words, $F_n$, does that work? – 9pound15pence Apr 19 '20 at 04:04
  • @9pound15pence I still don't see how these are equivalent. If $g_n(x) := \int_a^x f_n$ and similarly for $g$, then you don't know necessarily that $g_n \to g$ uniformly. You'd need to know that the integral of the $f_n$ on any _sub_interval approaches that of $f$. – Ethan Dlugie Apr 19 '20 at 04:08
  • @EthanDlugie Yes, That's my mistake, but does pointwise work? From condition (c), I write $\lim_{n\rightarrow\infty}(F_n(b)-F_n(a)) = F(b)-F(a)$ so $\lim_{n\rightarrow\infty} F_n(x) = F(x) \forall x \in [a,b]$. Is it correct? – 9pound15pence Apr 19 '20 at 04:20
  • @9pound15pence the same issue holds. Knowing what happens just at the ends doesn't tell you anything about what's happening in the middle. As a toy example, consider the constant sequence of functions $F_n(x)=sin(x)$ and $F(x) \equiv 0$ on $[a,b]=[0,\pi]$. Then $F_n(b)-F_n(a)=0=F(b)-F(a)$, but of course $F_n$ doesn't come close to $F$ anywhere in the middle. – Ethan Dlugie Apr 19 '20 at 04:25
  • @EthanDlugie Yes I see the problem here. And your hint in the answer is very helpful! – 9pound15pence Apr 19 '20 at 04:31

3 Answers3

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As another poster pointed out, there is a classic example that fulfills your request. But since this seems like a homework problem, it might be nice to see what the thought process might go like to get this yourself.

Without loss of generality, we can instead ask for a sequence $\{g_n\}$ of functions on $[a,b]$ such that each $g_n$ is integrable, $\lim_{n \to \infty} \int_a^b g_n =0$, but $\{g_n\}$ converges pointwise nowhere. There are basically two ways to make the integral go to zero:

(1) each $g_n$ is, say, positive but has 'mass' trending to zero, or

(2) each $g_n$ has positive and negative parts which cancel out in the integral.

Think about the ways you can get these conditions to happen but in a somehow incoherent way, so that for each $x \in [a,b]$ the sequence of values $g_n(x)$ does just bounces around without limit.

Ethan Dlugie
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  • So I come up with $f_n(x) = (-1)^{n+1}x-\frac{1}{2}$, then we have $f(x) =0$ fulfill condition (b) and (c). Moreover, ${f_n}$ diverges so satisfying condition (d) – 9pound15pence Apr 19 '20 at 04:30
  • What's the interval $[a,b]$ here? If you look on say $[-1,1]$, then $\int_{-1}^1 f_n = 0$ but $\lim_{n \to \infty} f_n(0) = 0$ exists. You need the function values to be bouncing around at every point. – Ethan Dlugie Apr 19 '20 at 04:32
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The typewriter sequence works. Take $f=0$ to get what you want.

Reveillark
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For $x \in [0,1]$ define $f_n(x) = (-1)^n([2x]-\frac{1}{2})$ (where $[\ ]$ is integer part), and $f(x)=0$

Sam
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