Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    I was given two questions, true and false, but am having trouble with them. They are...

    "if f is continuous at a, so is absolute value of f."

    and then
    "if absolute value of f is continuous at a, so is f."

    i am am at a loss at to how this works. The professor used a function for electric current as an example but It still seems to me that all points except zero would be continuous. Help!!!
    Offline

    3
    ReputationRep:
    (Original post by DottyPrawn)
    I was given two questions, true and false, but am having trouble with them. They are...

    "if f is continuous at a, so is absolute value of f."

    and then
    "if absolute value of f is continuous at a, so is f."

    i am am at a loss at to how this works. The professor used a function for electric current as an example but It still seems to me that all points except zero would be continuous. Help!!!
    1. Yes

    2, No
    • Thread Starter
    Offline

    0
    ReputationRep:
    Could you explain why? I feel like on the first one the continuous graph would still remain continuous if it was flipped over the x axis for negative numbers. On the second problem I think f(0) makes it false? I am very foggy on either problem.
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    (Original post by DottyPrawn)
    Could you explain why? I feel like on the first one the continuous graph would still remain continuous if it was flipped over the x axis for negative numbers. On the second problem I think f(0) makes it false? I am very foggy on either problem.
    First one: your intuition is sound. There's no way to become discontinuous without crossing the x-axis (as that's the only way to get the absolute function involved), and crossing the x-axis doesn't change continuity. Have you studied any analysis before? If so, you might well be able to come up with a proof yourself.

    Second one: we want a counterexample which is as simple as possible. What's the simplest discontinuous function you know of? Can you make that into an example of a continuous-in-absolute-value discontinuous function? (Depending on whether your idea of "simplest" is the same as mine, of course, you might not be able to, but do tell us anyway.)
    • Thread Starter
    Offline

    0
    ReputationRep:
    Would f(x) = -2 for x<0 and 2 for x > or = to zero work?

    So the absolute value would form a solid line at y=2 but f(0) would not be continuous because the limit from the negative would be -2 and the limit from the positive would be 2?
    Offline

    17
    ReputationRep:
    (Original post by DottyPrawn)
    Would f(x) = -2 for x<0 and 2 for x > or = to zero work?

    So the absolute value would form a solid line at y=2 but f(0) would not be continuous because the limit from the negative would be -2 and the limit from the positive would be 2?
    Yes, that works. For an even more striking example you can take Dirichlet's function

    f(x) = \begin{cases} 1, & \mbox{if } x\mbox{ is rational} \\ -1, & \mbox{if } x\mbox{ is irrational} \end{cases}

    Then |f(x)| = 1 for all x \in \mathbb{R} and so is continuous everywhere, yet f(x) is a nowhere continuous function (can you see why?)
    • Thread Starter
    Offline

    0
    ReputationRep:
    Would it be because without the absolute value there will always be going back and forth from 1 to -1 as you move across the x values?

    (thanks so much for the help too!)
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    (Original post by DottyPrawn)
    Would it be because without the absolute value there will always be going back and forth from 1 to -1 as you move across the x values?

    (thanks so much for the help too!)
    Yep, that's it - the irrationals and rationals are both dense in \mathbb{R}, so we can find an irrational and a rational arbitrarily close to any given point. Hence in any neighbourhood of a given point, the Dirichlet function takes both the values -1 and 1. (Alternatively, you could use the Intermediate Value Theorem: f(e) = -1 while f(3) = 1, so if f were continuous, f would have a root between e and 3. But f is nowhere 0 - it is only ever 1 or -1. Hence f is not continuous.)
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.