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August 16th, 2014, 06:09 PM   #1
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Convergent Sequence of Functions with Non-Zero Mean Square Error

I'm looking for an infinite sequence of functions, $f_n(x)$ that converge pointwise to zero on the closed interval $[0,1]$ but for which
$$\lim_{n \to \infty} \int_0^1 f_n^2(x) \, \mathbb{d}x \ne 0$$
Any ideas?

What about if the interval were open?
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August 17th, 2014, 08:25 AM   #2
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\begin{align*}f_n(x) &= \begin{cases} n & \tfrac{1}{n} \le x \le \tfrac{1}{n-1} \\ 0 &\text{otherwise} \end{cases} \\[12pt]
\lim_{n \to \infty} \int_0^1 f_n^2(x) \,\mathbb{d}x &= \lim_{n \to \infty} n^2\left( \frac{1}{n-1} - \frac{1}{n} \right) \\
&= \lim_{n \to \infty} n^2 \left( \frac{n - (n-1)}{n(n-1)} \right) \\
&= \lim_{n \to \infty} \frac{n^2}{n^2 - n} \\
&= \lim_{n \to \infty} \frac{1}{1 - \tfrac{1}{n}} \\
&= 1 \\
\end{align*}
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August 17th, 2014, 03:35 PM   #3
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Why not $\displaystyle f_{n}(x)=\begin{cases}n &,x\in\left(0,\frac{1}{n}\right)\\0 &,\text{otherwise}\end{cases}$?
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August 17th, 2014, 05:44 PM   #4
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It's not straightforward to show that your function tends to zero at each point of $[0,1]$. I'm tempted to say that for every $n$ there is a region $(0,a)$ for which $f_n(x) \ne 0$.

For any $c \in (0, 1]$ my sequence has $f_n(c) \ne 0$ for only one value of $n = N_c$. So for $n \gt N_c$ we have that $f_n(c) = 0$ and thus $f_n(x) \to 0$ for all points in $[0,1]$.

I suppose for your sequence you'd say that for each $c \in (0,1)$ there exists $N_c$ such that $f_n(c) = 0$ for all $n \gt N_c$, but there's still that region to the left that is non-zero. It's a conceptually more difficult solution.
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August 18th, 2014, 02:17 AM   #5
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$f_n(0)=0$ (we are in the "otherwise" category)
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