Prove that $\sqrt{x}\sin\frac{\pi}{2\sqrt{x}}$ is not of bounded variation and is not Lipschitz

Question

1) Prove that the function $f(x)$ is not of the bounded variation. 2) Prove that the function $f(x)$ is not Lipschitz continuous. $$f(x)=\left\{\begin{matrix} \sqrt{x}\sin\frac{\pi}{2\sqrt{x}} & \text{if} \ 0<x\leq1 \\ 0 & \text{if} \ x =0 \end{matrix}\right.$$

I can prove 1 easily. but don't know how to do 2. I thought it immediately follows from 1 but since it is a separate question I should be writing a proof for it.

let $x_i=\{(\frac{1}{2i+1})^2\}_{i\in N}$ , so \begin{align} TV(f,P) & = \sup_p \sum_{i=1}^n |f(x_i)-f(x_{i-1})|\geq\sum_{i=1}^n |f(x_i)-f(x_{i-1})|\\ & = \sum_{i=1}^n |(-1)^n (x_i+ x_{i-1})|=\sum_{i=1}^n x_i+ x_{i-1}\\ & = \sum_{i=1}^n\frac{1}{2i+1}+\frac{1}{2i-1} \to \infty\\ \end{align} so $f$ is not of the bounded variation. how do I show that is not liptschitz?

Answer 1

If it were Lipschitz, it would have bounded variation. So $1\implies2$.

Proof: Let $C$ be a Lipschitz constant for $f$. Then assuming $x_n$ is increasing with $n$, $$ \sup\sum_{i=1}^n|f(x_i)-f(x_{i-1})| \le C\sum_{i=1}^n|x_i-x_{i-1}|=C\sum_{i=1}^n(x_i-x_{i-1})=C(x_n-x_0)<\infty $$