I'm wondering if $$g \in L^2(\mathbb R)$$
then can I conclude that $$g(x) / (1 - x^2) \in L^2(\mathbb R)?$$
If $g \in L^2(\mathbb R)$ then can I conclude that $g(x) / (1 - x^2) \in L^2(\mathbb R)?$
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real-analysis
integration
functional-analysis
1 Answers
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No. Take $g$ to be the indicator of $[0,1]$. Then $\|g\|_2 = 1$, but
$$\int_{\Bbb R} \left(\frac{g(x)}{1-x^2}\right)^2\, dx = \int_{0}^1 \frac{1}{(1-x^2)^2}\, dx \ge \int_0^1 \frac{1}{4(1-x)^2}\, dx = \infty$$
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0Thanks but why the ineqaulity above holds? – 2017-01-15
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1@M.Doe since $x^2 - 2x + 1 \ge 0$, then $2 - 2x \ge 1 - x^2$, or $2(1 - x) \ge 1 - x^2$. Since $0 \le x \le 1$, then $[2(1 - x)]^2 \ge (1 - x^2)^2$, i.e., $4(1 - x)^2 \ge (1 - x^2)^2$. Hence $\frac{1}{(1-x^2)^2} \ge \frac{1}{4(1-x)^2}$ for all $x\in [0,1]$. – 2017-01-15