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Bounding the L-Infty Norm of a Diffble Fn

  1. Jul 27, 2011 #1
    Hello,

    I would appreciate any assistance with the following question: Suppose f \in C^2[-1,1] is twice continuously differentiable. Prove that

    |f'(0)|^2 \leq 4 ||f||_\infty (||f||_\infty + ||f''||_\infty), where ||f||_infty is the standard sup norm. At first I thought Taylor expansion seemed promising, but I am pretty stymied.

    Thank you!
     
  2. jcsd
  3. Jul 28, 2011 #2

    mathman

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    Fix your latex expression.
     
  4. Jul 28, 2011 #3
    Sorry, here's a fixed-up rewrite.
     
  5. Jul 31, 2011 #4
    First, as the inequality is invariant under the transformations [itex]f \mapsto cf[/itex], [itex]x \mapsto -x [/itex], we may assume WLOG that [itex]f(0), \ f'(0) \ge 0[/itex], and [itex]||f||_\infty = 1[/itex]. Thus, we must prove
    [tex]|f'(0)|^2 \le 4(1 + ||f''||_\infty)[/tex] subject to [itex] ||f||_\infty = 1[/itex]. Now, for all [itex]x \in (0, 1)[/itex] we have
    [tex] 1 \ge f(x) \ge f(0) + f'(0)x - \frac{x^2}{2} ||f''||_\infty, [/tex] so [tex] f'(0) \le \frac{1 - f(0)}{x} + \frac{x}{2} ||f''||_\infty.[/tex] The RHS is minimized for [itex]x^2 = x^2_0 = \frac{2(1 - f(0))}{||f''||_\infty}[/itex] and decreasing for [itex]0 < x < x_0[/itex]. If [itex]x_0 \le 1[/itex] then [tex] |f'(0)|^2 \le \frac{1}{x_0^2} \left( 1 - f(0) + \frac{x^2_0}{2} ||f''||_\infty \right)^2 = 2(1 - f(0))||f''||_\infty \le 4||f''||_\infty [/tex].
    If [itex] x_0 > 1[/itex] then [tex]|f'(0)|^2 \le \left(1 - f(0) + \frac{||f''||_\infty}{2} \right)^2 \le 4[/tex].

    In either case [itex]|f'(0)|^2 \le 4 + 4||f''||_\infty [/itex].
     
  6. Aug 2, 2011 #5
    namphcar,
    Thanks for the great response, it was quite helpful. Would you mind being a little more explicit in your last step?

    It's not clear to me that
    [tex]\left(1-f(0)+\frac{||f''||_{\infty}}{2}\right)^2\leq 4,[/tex]
    since a function of the form [itex]e^{kx}[/itex] would have an arbitrarily large second derivative.
    Thanks!
     
  7. Aug 2, 2011 #6
    Remember that [itex]x_0^2 = \frac{2(1 - f(0))}{||f''||_\infty}[/itex].
     
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