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Prove that $f:\mathbb{R}\to \mathbb{R}$ such that $$ f(x) = \left\{ \begin{array}{c l} x^2\, \sin\left(\frac{1}{x}\right) & ,\quad x\neq 0\\ 0 & ,\quad x=0 \end{array} \right.$$ is Lipschitz (without use of derivatives).

Attempt. I am aware (Lipschitz-continuous $f(x)=x^2\cdot \sin\left(\frac{1}{x}\right)$) that: $$|f(x)-f(y)|\leq 3|x-y| ~~~\forall~x,~y\in \mathbb{R},$$ but I am looking for a proof, without use of derivatives. I tried: for $x,y\neq 0$:

\begin{eqnarray} x^2\sin\frac{1}{x} - y^2 \sin\frac{1}{y} &=& (x^2-y^2)\sin\frac{1}{x} + y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right ),\nonumber \end{eqnarray} so: $$\left | x^2\sin\frac{1}{x} - y^2 \sin\frac{1}{y} \right | \leq |x^2 - y^2| + y^2 \left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |.$$ Since: $$\left | \sin\frac{1}{x} - \sin\frac{1}{y} \right | \leq \left| \frac{1}{x} - \frac{1}{y}\right |= \frac{|x - y|}{xy},$$ we get: $$\left | x^2 \sin\frac{1}{x} - y^2 \sin\frac{1}{y} \right | \leq \left(x+y+\frac{y}{x}\right)|x-y|.$$ Unfortunatelly , the quantity $x+y+\frac{y}{x}$ grows to $+\infty$, either for large $x$, or for $x\approx 0.$

Thanks in advance for the help.

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Writing $$ f(x) - f(y) = (x^2-y^2)\sin\frac{1}{x} + y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right ) $$ is a good start. The terms on the right-hand side can be estimated better if we assume that $0 < |y| \le |x|$: $$ \left| (x^2- y^2) \sin \frac 1x \right| \le |x-y| |x+y|| \frac 1x| \le 2|x-y| $$ and $$ \left|y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right )\right| \le |y^2| \left| \frac 1x - \frac 1y \right| = \left|\frac yx\right||x-y| \le |x-y| $$ and therefore $$ |f(x)-f(y)|\leq 3|x-y| \, . $$

For $0 < |x| \le |y|$ repeat the same calculation with $x$ and $y$ exchanged, or use the symmetry of $f$.

Finally, for $x \ne 0 = y$ $$ |f(x) - f(0)| = \left |x^2 \sin \frac 1x \right| \le |x| = |x-0| \, . $$

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    $\begingroup$ Great use of inequalities. Thank you! $\endgroup$ Commented Dec 16, 2018 at 22:38

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