March 16th, 2018, 07:12 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 07:59 AM. 
March 16th, 2018, 08:15 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,981 Thanks: 1853 
Yes. Does "integrable" in the question mean "Riemann integrable"?

March 16th, 2018, 09:16 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, 01:11 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 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, 04:17 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 520 Thanks: 293 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, 02:19 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,981 Thanks: 1853 
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?


Tags 
statement, true 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Is this statement true?  danielphillis  Applied Math  2  March 9th, 2014 07:06 PM 
true false for the statement about sifting  frankpupu  Linear Algebra  0  May 15th, 2012 01:35 PM 
Show the statement is true  Lucida  Algebra  10  March 26th, 2012 05:16 PM 
Is this statement about n distinct nonzero vectors true?  maximus101  Linear Algebra  3  February 22nd, 2011 06:12 AM 
Is this statement about n linearly dependent vectors true?  maximus101  Linear Algebra  1  February 17th, 2011 08:22 AM 