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 March 16th, 2018, 06:12 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Is it a true statement ? If $f: R -> R$ is integrable then $f(\sqrt {|x|})$ is integrable. Is it always true? Please explain. Thanks. Last edited by skipjack; March 16th, 2018 at 06:59 AM.
 March 16th, 2018, 07:15 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,190 Thanks: 1649 Yes. Does "integrable" in the question mean "Riemann integrable"? Thanks from Lalitha183
March 16th, 2018, 08:16 AM   #3
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Quote:
 Originally Posted by skipjack Yes. Does "integrable" in the question mean "Riemann integrable"?
The question is like this :
Pick up a true statement from the following, if $f: R -> R$
a) If $f$ is continuous then $|f|$ is continuous
b) If $f$ is differentiable then $|f|$ is differentiable
c) If $f$ is Integrable then $f(\sqrt {|x|})$ is Integrable
d) If $f$ is discontinuous then $|f|$ is discontinuous

This is Master's entrance question.

 March 16th, 2018, 12:11 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,331 Thanks: 2457 Math Focus: Mainly analysis and algebra a) is a true statement. I'd have thought c) was too. But if you only need one, I'd pick a). b) and d) are definitely false.
 March 16th, 2018, 03:17 PM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 398 Thanks: 212 Math Focus: Dynamical systems, analytic function theory, numerics (a) is the true statement. This is easily proved by considering an open subset of $\mathbb{R}$ and proving that its preimage is open under $|f|$. As a hint: consider the case that $0 \in f^{-1}(G)$ and $0 \notin f^{-1}(G)$ separately and use the fact that $[0,c)$ is an open subset of $\mathbb{R}^+$ for any $c$. (c) is definitely untrue. The problem here is that the function $y = \sqrt{x}$ does not have bounded variation. Thus, when you compute the differential of $f(\sqrt{x})$ it is clear that it is an unbounded linear operator. Specifically, suppose $g(x) = f(\sqrt{x})$, then its differential is given by $dg = \frac{f'(\sqrt{x}) \ dx}{2 \sqrt{x}}$ so that if $f'$ is continuous and $f'(0) \neq 0$, then $\left| \int_{0}^{\infty} dg \right| = \infty$. As a concrete example take $f(x) = 1$ for $x \in [0,1]$ and $f(x) = \frac{1}{x^2}$ for $x > 1$. Thanks from Lalitha183
 March 17th, 2018, 01:19 AM #6 Global Moderator   Joined: Dec 2006 Posts: 19,190 Thanks: 1649 For the purposes of the question, Lalitha183, were you expected to assume that "integrable" doesn't mean "Riemann-integrable", so that for a function to be "integrable" even an improper integral of it (such as an integral over an unbounded interval) must exist?

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