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I'm reading the book Real Analysis of Folland. When I reached chapter 2 about the different modes of convergence, there's an example Folland gave that confused me:

The 2 function sequences:
i) $f_n = \chi_{(n, n+1)}$
ii) $f_n = n\chi_{[0, 1/n]}$
uniformly, pointwise converge to $0$, but not converge in $L^1$

I can easily point out that those 2 functions pointwise converge to $0$ as well as not converge in $L^1$. But I don't think those 2 functions uniformly converge. Like the first case, for all $N$ and choose $\epsilon < 1$, we can always choose $x \in (N , N+1)$, then $|f_n(x)| = 1 > \epsilon$. Did I miss some points? (The example is in chapter 2, page 61) Please help me to judge this. Thanks so much.

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You misunderstood what he says. Folland writes:

(i) $f_n=n^{-1}\chi_{(0,n)}$

(ii) $f_n=\chi_{(n,n+1)}$

(iii) $f_n=n\chi_{[0,1/n]}$

In (i), (ii), and (iii), $f_n\to 0$ converges uniformly, pointwise, and a.e. respectively.

So $f_n=\chi_{(n,n+1)}$ does not converge uniformly, but pointwise which is evident. Only the first one converges uniformly.

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  • $\begingroup$ Ah yeah, so it means the first case converges uniformly, the second pointwise and the last a.e, right. Thanks so much for clarifying. $\endgroup$ – le duc quang Jul 23 '15 at 2:28

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