Is $f(x) = \sqrt x$ of bounded variation?

Question

I am trying to determine whether the $f(x) = \sqrt x$ is of bounded variation but only what I can proceed to is the definition of bounded variation:

$$V_a^bf=\sup\left\{\sum_{i=1}^{n}\left|\sqrt{x_i} - \sqrt {x_{i-1}}\right|, n \in \Bbb N, a=x_0<x_1<..<x_n=b\right\}$$

From above which step do I have to take to get to the conclusion of whether $f$ is of bounded variation or not?

$$V_a^bf=sup{\sum_{i=1}^{n}|\sqrt x_i - \sqrt {x_{i-1}}| n \in \Bbb N, a=x_0<x_1<..<x_n=b}=\ sup{\sum_{i=1}^{n}(\sqrt x_i - \sqrt {x_{i-1}}),\space n \in \Bbb N, a=x_0<x_1<..<x_n=b}$$ — Khosrotash, May 28 '17 at 05:52
It's of bounded variation on any bounded sub-interval of $[0,\infty)$. It is not of bounded variation of $[0,\infty)$, because it tends to infinity. — TonyK, Feb 22 '24 at 00:34

score 1 · Answer 1 · answered May 28 '17 at 06:00

$f(x)=\sqrt x$ is increasing function ,so $|\sqrt x_i - \sqrt {x_{i-1}}|=\sqrt x_i - \sqrt {x_{i-1}}$ then $$V_a^bf=\sup\left\{\sum_{i=1}^{n}\left|\sqrt{x_i} - \sqrt {x_{i-1}}\right|, n \in \Bbb N, a=x_0<x_1<..<x_n=b\right\}=\\ \sup\left\{\sum_{i=1}^{n}\left(\sqrt{x_i} - \sqrt {x_{i-1}}\right), n \in \Bbb N, a=x_0<x_1<..<x_n=b\right\}=\\ \sup\left\{(\sqrt{x_n}-\sqrt{x_{n-1}})+(\sqrt{x_{n-1}}-\sqrt{x_{n-2}})+(\sqrt{x_{n-2}}-\sqrt{x_{n-3}})+...+(\sqrt{x_2}-\sqrt{x_{1}})\right\}\\ \sup\left\{(\sqrt{x_n}-\sqrt{x_{1}})\right\}=\\b-a$$

Is $f(x) = \sqrt x$ of bounded variation?

1 Answers1