If $h:[a,b]\rightarrow\mathbb R$ Riemann integrable, then $H(x):=\int_a^xh(t)dt$ is Lipschitz continuous ($x\in[a,b]$).

Question

I have to solve the following problem:

Let $a,b\in\mathbb R$, $a<b$ and $h:[a,b]\rightarrow\mathbb R$ Riemann-integrable. Define $H:[a,b]\rightarrow\mathbb R$ by $H(x):=\int_a^xh(t)dt$. Prove that $H$ is Lipschitz continuous.

First of all, what does it mean for a function to be Riemann integrable? Does that mean it is Riemann integrable over every closed subinterval of its domain?

Second, I have no idea at all how I could approach this problem. Could anyone give a hint or explanation on how to solve this?

Answer 1

hint

$h $ integrable at $[a,b] \implies $

$h $ is bounded at $[a,b] \implies$

$ \exists M\in \mathbb R\; \ : \;\ \forall t\in [a,b] \; |h (t)| \le M $

$\implies \forall ( x,y)\in [a,b]^2 : y <x$

$ |H (x)-H (y)|=|\int_y^xh (t)|\le \int_y^x|h (t)|dt\le \int_y^xMdt =M (x-y)=M|x-y|.$

Answer 2

A positive real valued function is lipshitz continuous if

$\exists$ a real number $K >0$ such that for all real $x,y \quad |f(x)-f(y)|\leq K\cdot|x-y|$

Now, for $x,y \in [a,b]$ we have $$|H(x)-H(y)|= \bigg\vert \int_y^xh(t)dt\bigg\vert \leq \max_{x \in [a,b]}\{h(x)\}\cdot|x-y| $$ $\max_{x \in [a,b]}\{h(x)\}$is a positive constant because of the integrability of $h(x)$(Because if there is no such $M$ then $h$ won't have a convergent $U(P,f)$). Hence, $H(x)$ is lipshitz continuous.

A function is said to be Riemann integrable on a $\textbf{closed and bounded interval}$ if the upper and lower Riemann Sums of the function on that interval converge to the same value. Explained nicely here https://en.wikipedia.org/wiki/Riemann_integral