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I'm re-reading some material from Apostol's Calculus. He asks to prove that, if $f$ is such that, for any $x,y\in[a,b]$ we have

$|f(x)-f(y)|\leq|x-y|$

then:

$(i)$ $f$ is continuous in $[a,b]$

$(ii)$ For any $c$ in the interval,

$\left|\int_a^b f(x)dx-(b-a)f(c)\right|\leq\frac{(b-a)^2}{2}$

The proof for the first part is easy, and I ommit it. I'm interested in the second one.

We can write that as

$\left| {\int_a^b f (x)dx - \int_a^b f (c)dx} \right| \leqslant \frac{{{{(b - a)}^2}}}{2}$

Or $\left| {\int_a^b {\left( {f(x) - f(c)} \right)dx} } \right| \leqslant \frac{{{{(b - a)}^2}}}{2}$

Now, it is not hard to show that

$\left| {\int_a^b {\left( {f(x) - f(c)} \right)dx} } \right| \leqslant \int_a^b {\left| {f(x) - f(c)} \right|dx} $

By hypothesis, we have

$\left| {f(x) - f(c)} \right| \leqslant \left| {x - c} \right|$

so that

$\left| {\int_a^b {\left( {f(x) - f(c)} \right)dx} } \right| \leqslant \int_a^b {\left| {f(x) - f(c)} \right|dx} \leqslant \int\limits_a^b {\left| {x - c} \right|dx} $

The last term, integrates as follows:

$\int\limits_a^b {\left| {x - c} \right|dx} = - \int\limits_a^c {\left( {x - c} \right)dx} + \int\limits_c^b {\left( {x - c} \right)dx} = \frac{{{{\left( {b - c} \right)}^2} + {{\left( {a - c} \right)}^2}}}{2}$

How can I conciliate that with $\frac{{{{\left( {b - a} \right)}^2}}}{2}?$

I'd like to know what happens in the general case

$|f(x)-f(y)|\leq \lambda |x-y|$ too.

4 Answers 4

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Consider $f(c) = \dfrac{(b-c)^2+ (a-c)^2}{2}$. Then $f´(c) = (a-c)+(b-c)$. Hence, $f$ decreases of $a$ to $\dfrac{a+b}{2}$ and increases of $\dfrac{a+b}{2}$ to $b$. Hence $\dfrac{(a+b)^2}{2}=f(a)\ge f(c)$ to $a\le c\le \dfrac{a+b}{2}$ and $f(c) \le \dfrac{(a+b)^2}{2}=f(b)$ to $\dfrac{a+b}{2}\le c \le b$. Namely, $f(c) \le \dfrac{(a+b)^2}{2}$ to $c\in[a,b]$.

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You need to prove $ (b-c)^2 + (a-c)^2 \le (b-a)^2. $ A bit of algebra reduces this to $ c^2-bc-ac+ab \le 0. $ Factor by grouping: $ (c-a)(c-b)\le 0. $ This just says $c$ is between $a$ and $b$. It says that regardless of whether $a\le b$ or $b\le a$.

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    I'm not sure exactly what you mean here. If you mean $(a-b)^2/2$ rather than $(b-c)^2/2+(a-c)^2/2$, it could be that that is considered because it doesn't depend on $c$.2012-09-07
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Another way is to write $(b-a)^2= ((b-c)-(a-c))^2= (b-c)^2+ (a-c)^2-2(b-c)(a-c) \geq (b-c)^2+(a-c)^2$ because $(b-c)(a-c) \leq 0$.

For your second question, consider $\displaystyle \frac{1}{\lambda} f$ and apply your first result.

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I like pictures...

enter image description here

$\frac{(c-a)^2}{2} + \frac{(b-c)^2}{2} = m(T_1)+m(T_2) \leq \frac{m(R_1)+m(R_2)}{2} \leq \frac{1}{2} (b-a) \max(c-a,b-c) \leq \frac{1}{2} (b-a)^2$.

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    I don't understand. Do you mean why not use the formula with $c$? No reason, other than the homework...2012-09-07