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 $n$, $$ \sup\sum_{i=1}^n|f(x_i)-f(x_{i-1})| \le C\sum_{i=1}^n|x_i-x_{i-1}|=C\sum_{i=1}^n(x_i-x_{i-1})=C(x_n-x_0)<\infty $$