Is a continuous function bounded by Lipschitz continuous function also Lipschitz continuous?

Question

Is a continuous function bounded by Lipschitz continuous function also Lipschitz continuous?

For example this function

$$f(x,y)=\frac{x^3-3xy^2}{x^2+y^2}$$

and $f(0,0)=0$. I can show that

$$\left|f(x,y)\right| =\left|\frac{x^3}{x^2+y^2}-\frac{3xy^2}{x^2+y^2}\right| \leq\left|\frac{x^3}{x^2}\right|+\left|\frac{3xy^2}{y^2}\right| =4\left|x\right|$$

to prove $f(x,y)$ is continuous. However, since I know $4|x|$ is Lipschitz continuous. Can I also show that $f(x,y)$ is also Lipschitz continuous? Thank you.

Answer 1

No, this doesn't work as stated. $$f(x)=\begin{cases}0,\quad x<0\\1,\quad x\geq0\end{cases}\text{,}$$ the Heaviside step function, is trivially bounded by $1$ but not even continuous, let alone Lipschitz continuous.

Answer 2

Since $f$ is continuous at all points:

If the partials $\partial f/\partial x$ and $\partial f/\partial y$ exist and are continuous at all $(x,y)\ne (0,0),$ and if their absolute values for $(x,y)\ne (0,0)$ are bounded above by some $B\in \Bbb R^+$ then $f$ is Lipschitz:

Let $(x',y')=(x+d,y+e).$ Then $|f(x,y)-f(x+d,y)|\le B|d|$ and $|f(x+d,y)-f(x+d,y+e)|\le B|e|,$ so $$|f(x',y')-f(x,y)|\le B|d|+B|e|\le B\sqrt {d^2+e^2}=B\|(x',y')-(x,y)\|.$$

As a step in this direction, for the $f$ of the Q, if $(x,y)\ne (0,0)$ then $$|\partial f/\partial x|=|-x^4-6x^2y^2+3y^4|/(x^2+y^2)^2\le (3x^4+6x^2y^2+3y^4)/(x^2+y^2)^2=3.$$