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Real Analysis Mean Value Theorem Proof

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f: R->R be a function which satisfied f(0)=0 and |df/dx|≤ M. Prove that |f(x)|≤ M*|x|.


    2. Relevant equations
    Mean value theorem says that if f is continuous on [a,b] and differentiable on (a,b), then there is a point c such that f'(c)=[f(b)-f(a)]/(b-a).


    3. The attempt at a solution
    Let the derivative of f be between -M and M, and f(0)=0. For any point, p, I know that [f(p)-f(0)]/(p-0)= f(p)/p ≤ |M|.

    But I don't know where to go from here...
     
  2. jcsd
  3. Nov 1, 2012 #2

    jbunniii

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    Be careful with your absolute values. In fact, for any [itex]p \in \mathbb{R}[/itex], you know the following:
    [tex]\left|\frac{f(p) - f(0)}{p - 0}\right| = \left| \frac{f(p)}{p} \right| \leq M[/tex]
    However, you need to fill in the details to explain exactly how you are applying the MVT to deduce this inequality. This is the main part of the proof. Then, to finish the proof, all you have to do is recognize that
    [tex]\left| \frac{f(p)}{p} \right| = \frac{|f(p)|}{|p|}[/tex]
    and you're done.
     
  4. Nov 1, 2012 #3

    jbunniii

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    To apply the MVT, I suggest considering two cases: [itex]p > 0[/itex] and [itex]p < 0[/itex]. (The [itex]p = 0[/itex] case is obviously true without having to use the MVT.)

    For the first case, consider the interval [itex][0,p][/itex], and for the second case, consider [itex][p,0][/itex].
     
  5. Nov 1, 2012 #4
    Well the MVT says that if f is continuous on [a,b] and differentiable on (a,b), then there is a point c is f'(c)=(f(b)-f(a))/(b-a).

    Let's look at the interval [0,p] for a positive p. Since we're assuming f to be differentiable on the open region of this, we know there is some point such that f'(c)=f(p)/p, as said before. This slope, but assumption, can be no larger than M. So |f(p)/p| is less than or equal to M. This means that |f(p)|/|p| is less than or equal to M. Thus, |f(p)| is less than or equal to M*|p|... switch p to x, and I proved what was required.

    If we assume p to be negative, then look at [p,0]. Now by the MVT, we know there is some c such that f'(c)=(f(0)-f(p))/(0-p)= -f(p)/p. This must be less than or equal to M. So |f(p)/p|=|f(p)|/|p| is less than or equal to M as above.

    How do I get from that particular point c to a property that holds for the entire interval?
     
  6. Nov 1, 2012 #5

    LCKurtz

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    Maybe try an indirect argument -- what if there is a point p with |f(p)|>M|p|?
     
  7. Nov 1, 2012 #6
    Aha! Let's assume to the contrary that there IS a point, x, where |f(x)|>M*|x|. Then the MVT tells us that for the interval [0,x], there is a point c where the slope at c is equal to f(x)/x, but that is greater than M in absolute value terms, contradicting the assumption that |df/dx|< M. Similarly, if the point x were negative, I could make the same argument on the interval [x,0].

    Does that work?
     
  8. Nov 1, 2012 #7

    LCKurtz

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    That's the idea I was thinking. The details are yours...:smile:
     
  9. Nov 1, 2012 #8

    jbunniii

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    Exactly as you did above. What makes you think there's anything wrong with your proof? It looks fine to me. You might clean it up a bit by referring directly to x instead of p, such as the following:

    If [itex]x > 0[/itex], then we may apply the MVT to the interval [itex][0,x][/itex]. This tells us that there is some point [itex]c[/itex] such that [itex]0 < c < x[/itex] and
    [tex]f'(c) = \frac{f(x) - f(0)}{x - 0} = \frac{f(x)}{x}[/tex]
    Therefore,
    [tex]\left|\frac{f(x)}{x}\right| = |f'(c)| \leq M[/tex]
    so
    [tex]|f(x)| \leq M |x|[/tex]
    Note that the choice of [itex]c[/itex] depends on [itex]x[/itex], but all that matters is that for any [itex]x[/itex], we can find a [itex]c[/itex] that works.

    The proof is similar if [itex]x < 0[/itex]. Finally, if [itex]x = 0[/itex] then clearly [itex]0 = |f(x)| \leq M |x| = 0[/itex].
     
    Last edited: Nov 1, 2012
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