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Real analysis

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Mod note: Edited the function definition below to reflect the OP's intent.
    Suppose f:R->R is continuous. Let λ be a positive real number, and assume that for every x in R and a>0,f(ax)=aλ f(x). (a) If λ > 1 show that f is differentiable at 0. (b) If 0 < λ < 1 show that f is not differentiable at 0. (c) If λ=1, show that f is differentiable at 0 if and only if it is linear. (Hint: what is f(0)?)
    2. Relevant equations


    3. The attempt at a solution
    I am considering start the question with f(x)-f(0)/x but how can i find the limit when x approaches 0? f(x)=f(1*x)=1^λf(x) f(0)=f(0*x)=0^λf(x)=0, then is f(x)-f(0)/x = f(x)/x? Then how can i know it's differentiable at 0?
     
    Last edited by a moderator: Dec 6, 2015
  2. jcsd
  3. Dec 6, 2015 #2
    You must set up a limit, epsilon delta style
     
  4. Dec 6, 2015 #3
    how can i use epsilon delta to prove differentiability?
     
  5. Dec 6, 2015 #4

    Mark44

    Staff: Mentor

    You are missing parentheses. What you wrote is ##f(x) - \frac{f(0)}{x}##. I'm pretty sure you meant ##\frac{f(x) - f(0)}{x}##. If you don't use LaTeX like I did, you need to write it as (f(x) - f(0))/x.
    What does ^ mean here? It is often used to indicate exponentiation, as in 2^3 = 8.
    Are you using it to indicate "and"?
     
    Last edited: Dec 6, 2015
  6. Dec 6, 2015 #5
    You are right thats what i mean. yea ^ means exponentiation...
     
  7. Dec 6, 2015 #6
    "how can i use epsilon delta to prove differentiability?"

    Use the definition of the derivative.
     
  8. Dec 6, 2015 #7

    Mark44

    Staff: Mentor

    Then I'm really confused now. Where does this come from?
    I don't see anything in the problem statement about exponents.

    In any case, the above is very ambiguous. For example you have 1^λf(x), which I take to mean ##1^{\lambda}f(x)##, according to the normal rules for the order of operations. I don't know why you would have ##1^{\lambda}##, which equals 1, or ##0^{\lambda}##, which equals 0, as you have elsewhere.
     
  9. Dec 6, 2015 #8
    oh sorry i mistyped the problem. it should be f(ax)=a^lambda f(x) I think 1^lambda=1 because the problem states that lambda>1
     
  10. Dec 6, 2015 #9
    eq0002MP.gif ' of course the a in this definition is not the constant in the definition of f(x)
     
  11. Dec 6, 2015 #10

    Mark44

    Staff: Mentor

    Is it safe to interpret the above as ##f(ax) = a^{\lambda}f(x)##?
     
  12. Dec 6, 2015 #11
    yessss i just realize that f(0)=f(a*0)=a^lambda f(0), well then i totally get lost...
     
  13. Dec 6, 2015 #12
    yea thats what i think to start with but i can't find f(x) and f(0)
     
  14. Dec 6, 2015 #13

    Mark44

    Staff: Mentor

    The OP has to set up a limit, but I don't believe that a delta-epsilon argument is necessary.
     
  15. Dec 6, 2015 #14
    The nice thing about the definition of derivative AND f(x) is that you only need to find the difference, not the actual function.
     
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