Prove that $f: [a,b]\to \mathbb R$, $|f(x)-f(y)| \leq 4|x-y|$ is integrable, when $f$ reaches its maximum in any closed subinterval.

Question

Let $f:[a,b]\to \mathbb R$ such that it reaches its maximum and minimum in any closed sub-interval of $[a,b]$.

In addition:

$$|f(x)-f(y)| \leq 4|x-y|$$

holds for any $x,y$ in the interval. How would you prove that the function is integrable using the fact that Riemann's condition (i.e. $U-L < \epsilon$ for any epsilon)?

Answer 1

** HINT**

Let $\epsilon>0$ given.

take a subdivision $\sigma $ such that

its step $||\sigma||<\frac {\epsilon}{4 (b-a)} $.

in each subinterval $[x_i,x_{i+1}] $, we have

$$M_i-m_i=|f (\alpha_i)-f (\beta_i)|$$

$$\leq 4|\alpha_i-\beta_i|\leq 4||\sigma||$$

thus by summation,

$$U (f,\sigma)-L (f,\sigma)<\epsilon. $$