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The function $sin(x^2)$ is not Lipschitz continuous on $\mathbb{R}?$

My steps are like this:

Suppose assume for some $C\geq 0$ $$ \bigg|\frac{sin(x^2)-sin(y^2)}{y-x}\bigg|\leq C$$ is true, for all $y\neq x,$ where $x, y\in \mathbb{R}.$

Now how we should proceed further?

thomus
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    Compute the derivative $f'(x) = 2x\cos(x^2)$. Easy to verify $\limsup_{x \to +\infty} f'(x) = +\infty$ hence the derivative gets arbitrarily large. This implies that for any $M>0$ there are $x,y \in \mathbb{R}$ such that $\left|\frac{f(x) - f(y)}{x-y} \right|>M$, so $f$ is not Lipschitz. – MathematicsStudent1122 Jan 26 '18 at 06:46

1 Answers1

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It is even not uniformly continuous on the real line: Take $x_{n}=\sqrt{2n\pi+\pi/2}$ and $y_{n}=\sqrt{2n\pi}$, then $x_{n}-y_{n}\rightarrow 0$ but $\sin(x_{n}^{2})-\sin(y_{n}^{2})=1$ does not converge to $0$ as $n\rightarrow\infty$.

If it were Lipschitz continuous, then it is uniformly continuous.

user284331
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