# How to prove $f(x,y) = \frac{2 \log(y) \sqrt{y ( x-y)}}{x \log(x)}$ is bounded by $1$ for $(x,y) \to (0,0)$?

Let $$f(x,y) = \frac{2 \log(y) \sqrt{y ( x-y)}}{x \log(x)}$$ with $(x,y) \in D = \{(x,y)\mid 0 < y \leq x \le 1 \}$.

How would one show (or disprove) that $$\forall \epsilon > 0\ \ \exists \delta>0 \ \ \forall (x,y) \in D \cap B_\delta(0) \quad f(x,y) < 1 + \epsilon$$ where $B_\delta(0)$ is the disk with radius $\delta$ and center $(0,0)$?

Please note that this is not just a bound on the limit of the function, as it is not well-defined. Choosing another coordinate $0 \le \theta \le 1$ with $y = \theta x$ we get $$f(x, \theta x) = \frac{2 (\log(\theta) + \log(x)) \sqrt{\theta (1-\theta)}}{\log(x)}\\$$ such that $$\lim_{x \to 0} f(x, \theta x) = 2 \sqrt{\theta (1-\theta)}.$$ So the limit is not well-defined as it depends on how the origin is approached.

Also, although the above limit $2 \sqrt{\theta (1-\theta)}$ is indeed bound by 1, it does not directly prove the original question, because $\theta$ is assumed constant with respect to $x$. More specifically, this limit shows (variable ranges omitted for clarity) $$\forall \epsilon \ \ \forall \theta\ \ \exists \delta \ \ \forall x < \delta \quad f(x,\theta x) < 1 + \epsilon$$ while the original problem corresponds to the stronger (note the different quantifier ordering): $$\forall \epsilon \ \ \exists \delta \ \ \forall \theta\ \ \forall x < \delta \quad f(x,\theta x) < 1 + \epsilon$$

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In polar coordinates: $$f(r,\theta) = \frac{2 \log(r\sin \theta) \sqrt{\sin \theta (\cos \theta-\sin \theta)}}{\cos \theta \log(r\cos \theta)}$$

You can show that the function: $$2\sqrt{\sin \theta (\cos \theta-\sin \theta)} \sec (\theta) <1$$ Simply by taking the derivative and showing that it obtains the maximum value $1$ at $\theta\to\tan^{-1}(1/2)$. So: $$f(r,\theta)< \frac{\log(r\sin \theta)}{\log(r\cos \theta)} = \frac{\log(r)+\log(\sin \theta)}{\log(r)+\log(\cos \theta)}\sim 1+\frac{\sin \theta - \cos \theta}{\log(r)}+O(\log^{-2}(r))$$ Clearly, for small enough $r\to\delta>0$, $$\epsilon = \frac{\sin \theta - \cos \theta}{\log(\delta)} > 0$$.

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Thx for the interest! But I think the reasoning is invalid. As $$\frac{\log(r \sin \theta)}{\log(r \cos \theta)} = \frac{\log y}{\log x}$$ the expression can e.g. stay equal to 2 by choosing $y=x^2$ – veryltdbeard Feb 8 '13 at 17:53
@veryltdbeard - it can't, since $x>y$, and for small enough $r$ this is no longer true. – nbubis Feb 8 '13 at 18:47
As $x \le 1$, $x^2 \le x$, so choosing $y=x^2$ is valid. – veryltdbeard Feb 8 '13 at 20:37
I tried to add a clarification to the question pointing out the precise problem w.r.t taking limits. In case you're still interested, all help is still very welcome! – veryltdbeard Feb 12 '13 at 11:34

$$\begin{eqnarray} f(x, \theta x) &=& \frac{2 (\log(\theta) + \log(x)) \sqrt{\theta (1-\theta)}}{\log(x)}\\ &=& \frac{- 2 \log(\theta) \sqrt{\theta (1-\theta)}}{-\log(x)} + 2 \sqrt{\theta (1-\theta)} \\ &\le&\frac{- 2 \log(\theta) \sqrt{\theta (1-\theta)}}{- \log(x)} + 1 \\ &\le&\frac{V^*}{-\log(x)} + 1 \\ \end{eqnarray}$$ with $V^* = \max_{\theta \in [0,1]} (- 2 \log(\theta) \sqrt{\theta (1-\theta)}) \approx 1.3818$ (at $\theta^* \approx 0.1041$).

So $$\begin{eqnarray} f(x, \theta x) < 1 + \epsilon & \Leftarrow & \frac{V^*}{-\log(x)} + 1 < 1 + \epsilon \\&\Leftrightarrow & x < e^{-V^*/\epsilon} \end{eqnarray}$$ So with $\delta(\epsilon)=e^{-V^*/\epsilon}$ we have shown that $$\forall \epsilon > 0 \quad \forall x \in\ ]0, \delta(\epsilon)[\quad \forall \theta \in [0,1]\quad f(x, \theta x) < 1 + \epsilon.$$

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