Solve a problem

Statement of the problem:

f ' is Lipchitz, then

|\displaystyle\sum_{j=1}^n\frac{f_j+f_{j+1}}{2}-\int_0^1 f(x)\mathrm{d}x|\le CL\Delta x^2

for some constant C.

Proof:

We first state a lemma here.

f(x)-f(y)=f'(x)(x-y)+\displaystyle\frac{f'(\zeta)-f'(y)}{2(\zeta-y)}(x-y)^2,

for some \zeta\in(x,y).

Because we do not know whether f'' exists.

Proof for lemma:

Without loss of generality, we assume y=0.

Then we have to prove:

f(x)-f(0)=f'(0)x+\displaystyle\frac{f'(\zeta)-f'(0)}{2(\zeta-0)}x^2

Set g=f(x)-f'(0)x, then we have to prove:

g(x)-g(0)=\displaystyle\frac{g'(\zeta)}{2\zeta}x^2,

which is equivalent to:

\displaystyle\frac{g(x)-g(0)}{x^2}=\frac{g'(\zeta)}{2\zeta},

use the formula that, there exists a \phi, s.t.

\displaystyle\frac{F(x)-F(y)}{G(x)-G(y)}=\frac{F'(\phi)}{G'(\phi)}.

The lemma is proved.

Let’s prove the original problem. By the lemma,

f(x)-f_j=f'_j(x-j\Delta x)+\displaystyle\frac{f'(\zeta)-f'_j}{2(\zeta-j\Delta x)}(x-j\Delta x)^2,

f_{j+1}-f_j=f'_j\Delta x+\displaystyle\frac{f'(\eta)-f'_j}{2(\eta-j\Delta x)}\Delta x^2,

\displaystyle\int_{j\Delta x}^{(j+1)\Delta x}(f(x)-\displaystyle\frac{f_j+f_{j+1}}{2})\mathrm{d}x\le\int \{f'_j[x-(j+\frac{1}{2})\Delta x]+\frac{1}{2}L(x-j\Delta x)^2+\frac{1}{2}L\Delta x^2\}\mathrm{d}x

The first term is zero!

Then on every interval, the difference is of order three!

\Box

About YiMin

This is just a nerd PhD student of Math@UT Austin.