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Let $ f$ be a function such that $|f(u)-f(v)|\leq|u-v|$ for all real $u$ and $v$ in an interval $[a,b]$.Then:
$(i)$Prove that $f$ is continuous at each point of $[a,b]$.

$(ii)$Assume that $f$ is integrable on $[a,b]$.Prove that,$|\int_{a}^{b}f(x)dx-(b-a)f(c)|\leq\frac{(b-a)^2}{2}$,where $a\leq c \leq b$

I tried to solve second part,First part i could not get idea.
$|\int_{a}^{b}f(x)dx-(b-a)f(c)|=|\int_{a}^{b}f(x)-f(c)dx|=\int_{a}^{b}|f(x)-f(c)|dx\leq\int_{a}^{b}|x-c|dx\leq\int_{a}^{c}(c-x)dx+\int_{c}^{b}(x-c)dx$

But i am not getting desired result,what have i done wrong in this?Or is there another method to prove it.Please help.

diya
  • 3,589

2 Answers2

1

Hint:

$$(b-a)^2 \geqslant (b-c)^2 + (c-a)^2$$

RRL
  • 90,707
1

To prove the first assertion, let $\varepsilon > 0$ and let $c \in [a,b]$. Since $$ |f(x) - f(c)| \underset{\text{by assumption}}\leq |x-c| < \varepsilon $$ if $|x-c| < \varepsilon$, it follows that $f$ is continuous on $[a,b]$.

To prove the second assertion, note that, since $f$ is continuous on $[a,b]$, by the mean-value theorem for integrals there is some $c \in [a,b]$ such that $ \int_{a}^{b}f = f(c)(b-a), $ i.e. $\int_{a}^{b}f - [f(c)(b-a)] = 0 \leq (b-a)^{2}/2.$

I hope these are helpful.

Yes
  • 20,719