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Is is possible to show $f(z) \le 2 \sqrt{ \varepsilon} f(y)$ if $z < \varepsilon y$ for $f(x) = \sqrt{|x|\log \frac{1}{|x|}}$? I would like to use this inequality in a proof but I don't know wether it is true and how I can show it.

Any hints would be great!

  • 1
    Use the fact that $\log{x}$ is an ascending function. – 2017-01-03

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First of all, for $0This is the domain of $f(x)$.

Obviously $f(x) \geq 0$, $f(1)=f(-1)=0$. Also $f(x)=f(-x)$ as a result we can restrict the study to $0< x\leq 1$ (the other side is symmetric).

Function $f(x)$ has a maximum at $x=\frac{1}{e}$ and is ascending on $\left(0,\frac{1}{e}\right)$. This means that for $$0

However, if we still assume the inequality is true for the entire $0< x\leq 1$ :

  • For $y=1$ and $0only possible for $z=1$ - contradiction.
  • For $y=1-\frac{1}{n},n \in \mathbb{N}\setminus \{0,1\}$ and $z=\frac{1}{e}$ we can tune $0<\frac{1}{e}<\varepsilon \left(1-\frac{1}{n}\right)<1$, for example $\varepsilon=\frac{1}{2}$. Using this inequality $$f\left(\frac{1}{e}\right)=\frac{1}{\sqrt{e}}\le 2\sqrt{\varepsilon}f\left(y\right)=2\sqrt{\varepsilon}\sqrt{y\log{\frac{1}{y}}} \le 2\sqrt{\varepsilon}\sqrt{y\left(\frac{1}{y}-1\right)}=2\sqrt{\varepsilon}\sqrt{1-y}=2\sqrt{\frac{\varepsilon}{n}}$$ or simply $$\frac{1}{\sqrt{e}} \le 2\sqrt{\frac{\varepsilon}{n}}$$ which is wrong for large $n$ - another contradiction. In fact, an infinite sequence of contradictions if we note $y_n=1-\frac{1}{n}$.

So, there is no definitive answer as long as we keep the entire domain $0< |x|\leq 1$, further restrictions are required.

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    Thank you! That is a pity, that there is no definitive answer. So I can't use this inequality. But you assumed that $\varepsilon >1$, didn't you? So I think for $\varepsilon >0$ the whole problem becomes much more complicated, if not impossible. I will give it another try. Maybe I can give further restrictions. – 2017-01-03
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    For the 1st positive case on $(0,\frac{1}{e})$ - yes, $\varepsilon >1$. For the 2nd and 3rd negative cases though $\varepsilon < 1$. Yes, you could probably get a stable result by restricting further the domains of $f(x)$ and $\varepsilon$. – 2017-01-03
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    I found out that I can restrict $y < \varepsilon$. And then I can show that for $g(x):=f^2(x)=x \log \frac{1}{x}$ we have $g(z) < g(\varepsilon y) < 4 \epsilon g(y)$ which proves the statement. Thanks again for the help. – 2017-01-05