How to prove that $f: \mathbb{R} \to \mathbb{R}$ is continuous, given that $xy-y^2 \leq f(x)-f(y) \leq x^2-xy$?

Question

Let $f: \mathbb{R} \to \mathbb{R}$ and let $$xy-y^2 \leq f(x)-f(y) \leq x^2-xy$$ I am trying to prove that f is continuous i.e: $$\lim_{x \to x_0}f(x) = f(x_0)$$

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To be honest, I barely understand the problem. What I can extract from it is:

$$ xy-y^2 \leq x^2-xy \iff \boxed{x^2+y^2 \geq 1} $$

Hence, the inequality holds only $\forall$ outter points of the unit circle, centered at $(0,0)$.

More than that, I can't understand how to show continuity for $f$, since I don't even know what $f$ is. In order for a function to be continuous the following condition must be satisfied:

$$\lim_{x \to x_0} f(x) = f(x_0)$$

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Any tips that will help me understand the problem would be greatly appreciated

Answer 1

This follows from Squeeze Theorem. As $y \to x$, $x^{2}-xy$ and $xy-y^{2}$ both tend to $0$ and hence $f(x)-f(y) \to 0$. This means: $f$ is continuous at $x$.

[ For $x \neq 0$ check that $|x^{2}-xy| <\epsilon$ if $|x-y| <\frac {\epsilon} {|x|}$ and $|xy-y^{2}| <\epsilon$ if $|x-y| <\frac {\epsilon} {\epsilon+|x|}$. The case $x=0$ is trivial and leave it to you].

Answer 2

Hint: You need, for all $x\in\mathbb R$ and $\epsilon>0$, that there exists a $\delta>0$ so that $$|f(y)-f(x)|<\delta\text{ if }|y-x|<\epsilon.$$ This is a good definition of continuity to use, because it involves bounding the quantity $f(y)-f(x)$, which is exactly what you're given. In particular, you know that $$|f(y)-f(x)|\leq \max(|xy-y^2|,|x^2-xy|).$$ Can you show that, for all $x\in\mathbb R$ and $\epsilon>0$, that there exists a $\delta>0$ so that $$|xy-y^2|,|x^2-xy|<\delta?$$

Answer 3

Becuse $$0=\lim_{x\rightarrow x_0}(xx_0-x_0^2)\leq\lim_{x\rightarrow x_0}(f(x)-f(x_0))\leq\lim_{x\rightarrow x_0}(x^2-xx_0)=0.$$

Answer 4

Just to mention $$xy-y^2\le x^2-xy \iff x^2+y^2-2xy \ge 0 \iff (x-y)^2\ge 0$$ which is true for any $x,y\in \Bbb{R}$ so the inequality holds for all points in the plane, not only the outer points of the unit circle as you have mentioned.

Besides that since the given inequality holds for all pair of real numbers $x,y$, take $y=x_0$ and $\lim_{x\to x_0}$ on both sides of the given inequality $$\lim_{x\to x_0} (f(x)-f(x_0)) \ge \lim_{x\to x_0} (xx_0-x_0^2)=0 \\ \lim_{x\to x_0} (f(x)-f(x_0)) \le \lim_{x\to x_0} (x^2-xx_0)=0 \\ \implies \lim_{x\to x_0} (f(x)-f(x_0)) =0 $$

which is same as the squeeze theorem mentioned in other answers.