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On one paper I saw the function:

$$f_n=n_{\chi[0,1/n]}$$

What is this function?

I read that it's $n$ multiplied by the characteristic function on set $[0,1/n]$. But since $f_n$ in this case is supposed to defined for

$f \in L^1([0,1])$ s.t. $\int_0^1 f(t)dt=1$

then I don't see how $f_n$ necessarily stays bounded inside this.

mavavilj
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    have you computed $\int_0^1 f_n(t) dt$ ? – mercio Mar 05 '18 at 09:35
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    @mercio But what does $n_{\chi[0,1/n]}$ mean? Does it mean that $\chi$ modulates $n$ so that $n$ appears only when $\chi$ gets value $1$ or what? Then the integral is probably equal to $|_0^1 \frac{1}{2}t^2=\frac{1}{2}(1^2-0^2)=\frac{1}{2}$? – mavavilj Mar 05 '18 at 09:39
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    It is $n$ multiplied with the characteristic function of $[0, \frac 1n]$. Though are you sure the $\chi$ is a subscript of $n$ ?? – mercio Mar 05 '18 at 09:50
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    @mercio What does the characteristic function characterize? Since it's "if $x \in A$, then $1$, otherwise $0$", but what's $x$? – mavavilj Mar 05 '18 at 09:51
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    $\chi_A(x) = 1$ if $x \in A$ and $\chi_A(x) = 0$ if $x \notin A$ – mercio Mar 05 '18 at 10:26

1 Answers1

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Wait, there's a misunderstanding here!

Let: $$f_n:=n\chi_{[0,1/n]}$$ and also let: $$f:=\lim_{n\to\infty}f_n$$ Then, it is clear that: $$I_n:=\int_0^1f_nd\lambda=\int_0^{1/n}ndt=1$$ for every $n\in\mathbb{N}$. So, every $f_n$ seperately is bounded.

However, the sequence $\{f_n\}$ is not uniformly bounded!. Moreover, $$f(x)=\left\{\begin{array}{ll} \infty & x=0\\ 0 & x\neq0 \end{array}\right.$$ which is clearly not bounded.

Also note that: $$\int fd\lambda=0$$

Vassilis Markos
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