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: 19,540 Thanks: 1750 
Yes. Does "integrable" in the question mean "Riemann integrable"?

March 16th, 2018, 08:16 AM  #3  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
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: 7,403 Thanks: 2477 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: 444 Thanks: 254 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 $x > 1$. 
March 17th, 2018, 01:19 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,540 Thanks: 1750 
For the purposes of the question, Lalitha183, were you expected to assume that "integrable" doesn't mean "Riemannintegrable", 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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