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Let $V$ be a neighborhood of the origin in ${\Bbb R}^2$ and $f:V\to{\Bbb R}$ be continuously differentiable. Assume that $f(0,0)=0$ and $f(x,y)\geq -3x+4y$ for $(x,y)\in V$. Prove that there is a neighborhood $U$ of the origin in ${\Bbb R}^2$ and a positive number $\epsilon$ such that, if $(x_1,y_1),(x_2,y_2)\in U$ and $f(x_1,y_1)=f(x_2,y_2)=0$, then $$ |y_1-y_2|\geq\epsilon|x_1-x_2|. $$

Using the assumption, we have $$ f(x)=f'(0)x+o(\|x\|) $$ which gives the local behavior of $f$ near the origin. But how the inequality $f(x,y)\geq -3x+4y$ would be used here?

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The inequality means that $f$ always lies above a certain plane determined by the value of the linear combination. Since $f$ is differentiable, it has a tangent plane at the origin, so if the two planes are not the same, you can find a suitable point close to the origin where $f$ is asymptotically close to the second plane and hence lies below the first plane, and hence a contradiction.

user21820
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  • How does this give the inequality $|y-y'|\geq \epsilon |x-x'|$? Lying above a plane does not preempt having $y=y'$. – Behnam Esmayli Jul 24 '20 at 21:22
  • @BehnamEsmayli: I didn't say what I thought I did. I said "if the two planes are not the same, ...". Once you eliminate that case, then you would have found the tangent plane, which implies that near the origin the curve of the equation $f(x,y) = 0$ is approximately a certain line, which yields the desired inequality. – user21820 Jul 25 '20 at 04:24
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Note that $f_y(0,0) \geq 4,$ and hence $Df (0,0) \neq 0.$ Hence the implicit function theorem tells you that that there is a neighborhood $(-\epsilon,\epsilon)$ of $0 \in \mathbb{R}$ and a $C^1$ diffeomorphism $g: (-\epsilon, \epsilon) \to g (-\epsilon, \epsilon)$ such that $f(x,y) = 0 \Rightarrow y = g(x),$ for all appropriate $x,y.$

Hence given $(x_1,y_1)$ and $(x_2,y_2)$ with $x_1 \neq x_2$ and those points sufficiently close to $0,$ and satisfy $f(x_j,y_j) = 0, j = 1,2$ then we see $y_1 \neq y_2$ (by the local injectivity part of the implicit function theorem). Hence $y_1 - y_2 = g(x_1) - g(x_2) = g'(\xi) (x_1 - x_2),$ for some point $\xi \in (x_1,x_2).$

Then note that $g' \neq 0$ on $(-\epsilon,\epsilon).$ The conclusion follows since $g$ is $C^1.$ ($g'$ has a positive minimum on a compact subset of $(-\epsilon,\epsilon)$ containing $x_1,x_2.$ )

Raghav
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