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Finding a Maximum

  1. Nov 15, 2008 #1
    The problem statement, all variables and given/known data
    Let f: U -> R where U is an open subset of R. Suppose that for some c in U, f'(c) = 0 and f''(c) < 0. Proof that the restriction of f to some open ball of center c attains a maximum at c.

    The attempt at a solution
    Let e > 0. Then for some d' > 0, if |x - c| < d' then |f(x) - f(c)| ≤ e|x - c| since f'(c) = 0. Let f''(c) = -b for some positive b. Then for some d'' > 0, if |x - c| < d'' then |f'(x) + b(x - c)| ≤ e|x - c|. Thus, for d = min{d', d''}, if |x - c| < d then

    |f(x) - f(c)| ≤ e|x - c| and |f'(x) + b(x - c)| ≤ e|x - c|.

    Of course, what I need is f(x) ≤ f(c). I don't see how to get that from the above. Any tips?
  2. jcsd
  3. Nov 15, 2008 #2


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    Assuming you know f'(x) > 0 implies f is increasing in a ball around x:

    Sketch argument:
    f''(c)<0 means f' is decreasing in a ball around c. Hence as f'(c)=0, f'(x)>0 if x<c and f'(x)<0 if x>c for x close to c. If c isn't a maximum, there exists x s.t. f(x)>f(c). -> in between x and c, there exists d s.t. f'(d) = f(x)-f(c)/(x-c) What can you say about the sign of f'(d)?
  4. Nov 15, 2008 #3
    I was not aware of this fact. I will attempt to prove it on my own.

    If f'(d) is negative, x < c, but then we get a contradiction since f'(x) > 0 if x < c. A similar contradiction occurs if f'(d) is positive. So then f'(d) = 0 and f(x) = f(c), another contradiction. Thus c is a maximum.

    Cool. Thanks a lot.
  5. Nov 15, 2008 #4
    This is more difficult than I thought. All I've been able to derive is the following: Let e > 0. Then there is a d > 0 such that if |y - x| < d, then

    |f(y)| < (e + f'(x))|y - x| + |f(x)|.

    But this doesn't help in determining that f(a) ≤ f(b) if a < b where a, b are in the ball of radius d centered at x.
  6. Nov 15, 2008 #5


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    If f'(y)>0 for some y, then [tex]\lim_{x \rightarrow y} \frac{f(x)-f(y)}{x-y} > 0[/tex]

    So heuristically (you can make this rigorous), near y, if x<y, f(x)<f(y) as otherwise the [tex]\frac{f(x)-f(y)}{x-y} < 0[/tex] and if x>y, f(x)>f(y) as otherwise [tex]\frac{f(x)-f(y)}{x-y} < 0[/tex] just by noting the numerator and denominator have opposite signs
  7. Nov 15, 2008 #6
    It never occured to me to look at [f(x) - f(y)] / (x - y). I understand your idea. However, that only shows that for some ball B centered at y, if x in B and x < y, then f(x) < f(y). It doesn't prove that f is increasing in B. For example, let x' in B and suppose x < x' < y. What can we conclude about f(x) versus f(x')?
  8. Nov 16, 2008 #7
    I don't think we need for f to be increasing. It's enough to show that f'(x)>0 if x<c and f'(x)<0 if x>c for x close to c.

    Another comment: You seem to be using the mean value theorem for the existence of d, but we can't use it for we don't know whether f is continuous and differentiable on [x,c] (or [c,x]).
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