I'm facing some difficulty in proving/disproving this sentence:
Consider $f: [0, \infty ) \rightarrow \mathbb{R}, f$ is continuous.
if $\displaystyle \int_1^\infty f(x)dx$ exist, then $\displaystyle \int_1^\infty f^2(x)dx$ exist.
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user2637293
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1what do you think? what have you tried? – Lost1 May 14 '14 at 11:35
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Do you have any knowledge of $ L^p $ spaces? – derpy May 14 '14 at 11:36
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^ complete unnecessary, the proof is an introductory analysis one.... – Lost1 May 14 '14 at 11:36
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@Lost1 Never said it was necessary, mate... – derpy May 14 '14 at 11:37
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@Lost1 i think it does, but i don't know how to prove this. – user2637293 May 14 '14 at 11:42
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2Try something like $f(x) = \sin(x)/\sqrt{1 + x}$. – Tunococ May 14 '14 at 11:45
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@Tunococ how did you show that your example converge? – user2637293 May 14 '14 at 12:06
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1In fact we have $f(x)=e^{-|x|}/\sqrt{|x|}\in L^1\setminus L^2$ and $g(x)=1_{[1,\infty)}(|x|)/|x|\in L^2\setminus L^1$. – May 14 '14 at 12:28
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@Edwin: your first example does not work in this case, the OP asks for integration over $[1,\infty)$. – Clement C. May 14 '14 at 13:24
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@ClementC. You are right. Thank you for pointing out my typo. – May 14 '14 at 14:52
2 Answers
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Classic example: define $f$ to be piecewise affine and made of disjoint isocele triangles ("bumps"; and $0$ elsewhere), the $n$-th of them, centered at $x_n=n$, having base of length $2/n^3$ and height $n$.
Clement C.
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I'll just expand my comment here. One thing I want to point out is that Clement's example is, in a sense, better because it is unconditionally integrable while my example $f(x) = (\sin x) / \sqrt{1 + x}$ is conditional.
To show that my function is conditionally integrable, I use the definition $\int_1^\infty f(x) dx = \lim_{y \to \infty} \int_1^y f(x) dx$. In fact, I'll show that $\int_0^\infty f(x) dx = \lim_{y \to \infty} \int_0^y f(x) dx$ exists.
First, I'll prove that $\lim_{n \to \infty} \int_0^{2n\pi} f(x) dx$ exists. For each non-negative integer $n$, define $s_n = \int_{2n\pi}^{2(n+1)\pi} f(x) dx$. We can bound each $s_n$ as follows. \begin{align} p_n = \int_{2n\pi}^{(2n+1)\pi} f(x) dx & = \int_{2n\pi}^{(2n+1)\pi} \frac{\sin x}{\sqrt{1 + x}} dx \\ & \le \int_{2n\pi}^{(2n+1)\pi} \frac{\sin x}{\sqrt{1 + 2n\pi}} dx \\ & = \frac{2}{\sqrt{1 + 2n\pi}} \\ m_n = \int_{(2n+1)\pi}^{(2n+2)\pi} f(x) dx & = \int_{(2n+1)\pi}^{(2n+2)\pi} \frac{\sin x}{\sqrt{1 + x}} dx \\ & \le \int_{(2n+1)\pi}^{(2n+2)\pi} \frac{\sin x}{\sqrt{1 + (2n + 2)\pi}} dx \\ & = -\frac{2}{\sqrt{1 + 2(n + 1)\pi}} \end{align} Therefore, \begin{align} s_n = p_n + m_n & \le \frac{2}{\sqrt{1 + 2n\pi}} - \frac{2}{\sqrt{1 + 2(n+1)\pi}}. \end{align} It is easy to see that $s_n \ge 0$, so the sequence $\{S_n\}$ of partial sums $S_n = \sum_{i=0}^n s_i$ is monotonically increasing and bounded above by $2$ (the sum is telescoping), hence convergent.
For $y \in [2n\pi, (2n+1)\pi]$, we have $S_n \le \int_0^y f(x) dx \le S_n + p_n$. For $y \in [(2n+1)\pi, 2(n+1)\pi]$, we have $S_n + p_n \ge \int_0^y f(x) dx \ge S_{n+1}$. This is sufficient to conclude that $\int_0^\infty f(x) dx = \lim_{y \to \infty} \int_0^y f(x) dx = \lim_{n\to\infty} S_n$.
Tunococ
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