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If $f(x)=\frac{\cos(x)-\cos(a)}{x-a}$, where $a$ is a fixed number, how to prove the following inequality \begin{equation} |f(x_1)-f(x_2)|\leq C|x_1-x_2|,\quad \forall~~ x_1,x_2\in \mathrm{R}.~~~~~~~~~~(1) \end{equation}

In particular, let $a=0$, we have $f(x)=\frac{\cos(x)-1}{x}$, I've plotted the graphic of $f^{\prime}(x)$ with $x\in[-20\pi,20\pi]$ by using MATLAB:

enter image description here

It seems that $f^{\prime}(x)$ is bounded and the result $(1)$ holds. However, could anyone show me how to prove this conclusion $(1)$?

The MATLAB code is as follows:

x=-20*pi:0.003*pi:20*pi;
y1=(-x.*sin(x)-cos(x)+1)./(x.^2);
plot(x,y1)
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  • $\begingroup$ Hint: check mean value theorem. This looks like a textbook problem/assignment so I am going to give a full answer, but the hint should be more than enough for you to figure out the proof yourself. $\endgroup$ – Abdullah Ali Sivas Apr 5 at 13:35
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    $\begingroup$ This is a math question. You should ask it on the Math StackExchange forum. $\endgroup$ – Wolfgang Bangerth Apr 6 at 2:37
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In terms of divided differences, you have $f(x)=g[x,a]$, $g(x)=\cos(x)$. Then the divided difference of $f$ is, by the recursive definition of them and their symmetry, the order 2 divided difference of $g$, $$ f[x,y]=\frac{g[a,y]-g[x,a]}{y-x}=g[x,a,y]=g[a,x,y] $$ It is known that there is some $\xi$ in the interval containing $a,x,y$ so that $$ g[a,x,y]=\frac12g''(\xi) $$ If $g''$ is bounded, as here with $g''(x)=-\cos(x)$ by $1$, then this bound, divided by 2, is a Lipschitz constant for $f$.

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