Monday, 2 December 2013

calculus - Proving Schwarz derivative fracf(0)2=limlimitsxto0fracfracf(x)f(0)xf(0)x without Taylor expansion or L'Hopital rule?



Assume that f is C2 near 0. I would like to show the following Schwartz derivative
f(0)2=limx0f(x)f(0)xf(0)x








I am able to do this by using the Taylor expansion and L'Hopital rule. I am wondering how one can prove it without using Taylor expansion or L'Hopital rule.



Answer



Suppose x>0. Note
\begin{eqnarray} &&\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}\\ &=&\frac{f(x)-f(0)-xf'(0)}{x^2} \\ &=&\frac{\int_0^xf'(t)dt-\int_0^xf'(0)dt}{x^2} \\ &=&\frac{\int_0^x[f'(t)-f'(0)]dt}{x^2} \\ &=&\frac{\int_0^x\bigg[\int_0^tf''(s)ds\bigg]dt}{x^2} \\ &=&\frac{\int_0^x\bigg[\int_s^xf''(s)dt\bigg]ds}{x^2} \\ &=&\frac{\int_0^x(x-s)f''(s)ds}{x^2} \end{eqnarray}
and
\int_0^x(x-s)ds=\frac12x^2.
So
\begin{eqnarray} &&\bigg|\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}-\frac12f''(0)\bigg|\\ &=&\bigg|\frac{\int_0^x(x-s)[f''(s)-f''(0)]ds}{x^2}\bigg|\\ &\le&\bigg|\frac{\int_0^x(x-s)|f''(s)-f''(0)|ds}{x^2}\bigg| \end{eqnarray}
Since f\in C^2, for \forall \varepsilon>0, \exists \delta>0 such that
|f''(x)-f''(0)|<\varepsilon \forall x\in(0,\delta).
Thus for x\in(0,\delta),
\begin{eqnarray} &&\bigg|\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}-\frac12f''(0)\bigg|\\ &\le&\frac{\int_0^x(x-s)|f''(s)-f''(0)|ds}{x^2}\\ &\le&\bigg|\frac{\int_0^x(x-s)\varepsilon ds}{x^2}\bigg|\\ &=&\frac12\varepsilon. \end{eqnarray}
So
\lim_{x\to0}\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}=\frac12f''(0).


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