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I have two parameters $\alpha,\varepsilon>0$ and the following difference: $$D:=\left|\,\varphi\left(\frac{x-\alpha^2-\varepsilon}{\alpha}\right)-\varphi\left(\frac{x-\alpha^2+\varepsilon}{\alpha}\right)\,\right|,$$ where $\varphi(z)=\frac{1}{2\pi}e^{-z^2/2}$. I have the impression that $D\leq C\varepsilon$ for some constant $C>0$ independent of $x$ and $\alpha$. Nevertheless, applying the Lipschitz property "brutaly" we get $$D\leq \max_{z}\left|\varphi'(z)\right| \frac{2\varepsilon}{\alpha} \leq C \frac{\varepsilon}{\alpha}.$$

Observe that when we send $\alpha\to 0$ then the RHS above explodes, while in $D$ it seems to be bounded. Where is the error? or how can I show that the above estimate holds true? Any idea?

Thanks a lot!

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  • $\begingroup$ Why do you think there is an error? $\endgroup$ – zhw. May 16 '16 at 17:52
  • $\begingroup$ Well, no but I always thought the Lipschitz property was quite sharp... But if it's true I don't manage to prove it (nor disprove it). Actually I had an idea that seems to work :) but I would like to confirm that it is true... $\endgroup$ – Martingalo May 16 '16 at 18:46
  • $\begingroup$ Take $\alpha = \epsilon^2$ and fix $x = \epsilon^4 + \epsilon$ then $$D = \frac{1}{2\pi}\left[e^{-\frac{1}{2\epsilon^2}} - 1\right] \to \frac{1}{2\pi}$$ Thus you cannot have $D \leq C\epsilon$ with $C$ independent of $x$ and $\alpha$. $\endgroup$ – Winther May 20 '16 at 14:38
  • $\begingroup$ what does RHS mean? $\endgroup$ – moonshine May 26 '16 at 18:39
  • $\begingroup$ Right-hand side. $\endgroup$ – Martingalo May 27 '16 at 11:12
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Given any $\varepsilon\gt0$, let $x=\alpha^2+\epsilon$. Then $\frac{x-\alpha^2-\varepsilon}{\alpha}=0$ and $\frac{x-\alpha^2+\varepsilon}{\alpha}=\frac{2\varepsilon}\alpha$. Then $$ \varphi\left(\frac{x-\alpha^2-\varepsilon}{\alpha}\right)-\varphi\left(\frac{x-\alpha^2+\varepsilon}{\alpha}\right)=\frac1{2\pi}-\varphi\left(\frac{2\varepsilon}{\alpha}\right)\tag{1} $$ And by letting $\alpha\to0$, we get that the difference in $(1)$ tends to $\frac1{2\pi}$ no matter how small $\varepsilon$ is. Therefore, it is not possible to get $$ \left|\,\varphi\left(\frac{x-\alpha^2-\varepsilon}{\alpha}\right)-\varphi\left(\frac{x-\alpha^2+\varepsilon}{\alpha}\right)\,\right|\le C\varepsilon\tag{2} $$ independently of $x$ and $\alpha$.

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  • $\begingroup$ Aha, but is it possible to get dependence on $\alpha$ which is integrable at 0, that means better than $\frac{1}{\alpha}$? like for instance $\frac{1}{\sqrt{\alpha}}$ or similar? $\endgroup$ – Martingalo May 21 '16 at 11:50
  • $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn May 25 '16 at 2:05
  • $\begingroup$ Thanks! that's what I expected :) thank you. $\endgroup$ – Martingalo May 25 '16 at 10:05
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It's an inequality. You may have given something away on the right hand side, but that doesn't imply an error.

A better estimate would be

$$|D| \le \sup \{|\varphi '(z)|: (x-\alpha^2 -\epsilon)/\alpha \le z \le (x-\alpha^2 +\epsilon)/\alpha\}\cdot (2\epsilon/\alpha).$$

Since $|\varphi '(z)| \to 0$ fast as $|z|\to \infty,$ the above should help.

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  • $\begingroup$ How do you get this estimate? Why do you take the maximum over a shorter interval? and how do you get rid of the $\alpha$ dividing? :) thanks. $\endgroup$ – Martingalo May 16 '16 at 19:23
  • $\begingroup$ It's just the mean value theorem, the same thing you used in your first estimate. $\endgroup$ – zhw. May 24 '16 at 21:43

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