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Thread: tayler's theorem

  1. #1
    Junior Member
    Mar 2009

    tayler's theorem

    Let f : V W be C2, where V and W are normed vector spaces. Define

    M_0 = sup{f(x)∥_w : x V}

    M_1 = sup{f(x)∥_L(v,w) : xV}

    M_2 = sup{f(x)∥_L2(v,w) : xV}

    Assume that these are all finite. Show that

    M_1^2 ≤ 4M_0M_2.

    [Hint: use Taylors theorem to show that for any xV and vV, and any s>0,

    f(x)(v)W ≤ (f(x+sv) f(x)W)/s + (s/2)M_2,

    And so deduce f(x)L(v,w) ≤ 2M_0/s + sM_2/2. Choose s carefully.]

    could anyone please help me with this problem ??
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    It will help if you understand the scalar case before attempting the vector problem. Suppose that f:\mathbb{R}\to\mathbb{R} is a C^2-function. Let M_0 = \sup|f(x)|, M_1 = \sup|f'(x)|, M_2 = \sup|f''(x)|, supposing that these are all finite. Taylor's theorem says that for any x\in\mathbb{R} and s>0, f(x+s) = f(x) + sf'(x) + \tfrac12s^2f''(\xi) for some \xi between x and x+s. Therefore f'(x) = \tfrac1s\bigl(f(x+s)-f(x)\bigr) - \tfrac s2f''(\xi). By the triangle inequality, |f'(x)| \leqslant \tfrac1s|f(x+s)-f(x)| + \tfrac12s|f''(\xi)| \leqslant \tfrac2sM_0 + \tfrac s2M_2.

    Now let s = 2\sqrt{M_0/M_2}. The previous inequality becomes |f'(x)|\leqslant 2\sqrt{M_0M_2}. Square both sides and take the sup over x to get M_1^2\leqslant 4M_0M_2.

    That proves the scalar case. Now you have to carry essentially the same argument over to the case of vector-valued mappings.
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