I'm trying to prove the following inequality for $0 < x < 1$:

$$\operatorname{erf}\left(\frac{(1+x)\sqrt{\ln{(1+x)}}}{\sqrt{(1+x)^2 - 1}}\right) - \operatorname{erf}\left(\frac{\sqrt{\ln{(1+x)}}}{\sqrt{(1+x)^2 - 1}}\right) \geq \frac{x}{4}$$

Proof by WolframAlpha: http://goo.gl/15mrM

I could also construct a Proof by Mathematica, without too much trouble.

However, I'm looking for a more elegant proof of this inequality. My approach was going to involve showing that this holds for $x = 0$ and $x = 1$, and then show the function is concave. However, taking the second derivative yields the following monstrosity: http://goo.gl/fKxca

Is there a more elegant way to prove this? I wouldn't mind showing a weaker inequality of the form $\geq \frac{x}{c}$ (for some explicit $c$) if the proof was sufficiently simple.

  • $\begingroup$ I agree that erf(x) is concave. However, the difference of two concave functions is not necessarily concave. Consider the first function as y = -0.5x^2, and the other as y = -x^2. The difference will be y = 0.5x^2, which is convex. $\endgroup$ – Gautam Kamath Jul 17 '13 at 2:57
  • $\begingroup$ @newzad: as Gautam pointed out, it isn't because $f, g$ are concave that the difference of the two is (in other terms for $\mathcal ¢^2$ functions, $f'', g'' \leq 0 \not\Rightarrow f''-g'' \leq 0$, clearly). Similarly, the problem is not erf, but the error function composed with another function; and again, in general, the concavity of $f\circ g$ does not follow from the concavity of $f$ (or even of $f$ and $g$). $\endgroup$ – Clement C. Jul 17 '13 at 9:26

I don't know if it's the elegant approach you're looking for, but here's a suggestion: fix any $x_0 > $, and define $$ f_{x_0}\colon y > 0 \mapsto \operatorname{erf}\left(\frac{(1+y)\sqrt{\ln(1+x_0)}}{\sqrt{(1+x_0)^2-1}}\right) $$ Now, what you want to prove is $$\begin{equation} f_{x_0}\!(x_0)-f_{x_0}\!(0) \geq \frac{x_0}{4} \tag{1} \end{equation}$$ so it is sufficient to prove that for all $y\in[0,2x_0]$, $$\begin{equation} f_{x_0}\!(y)-f_{x_0}\!(0) \geq \frac{y}{4} \end{equation}$$ i.e. $$\begin{equation} \frac{f_{x_0}\!(y)-f_{x_0}\!(0)}{y} \geq \frac{1}{4} \tag{2} \end{equation}$$ Since $f_{x_0}$ is concave, you can use the usual arguments about concavity/convexity (eg, a concave function has a decreasing slope).

Does that make sense? (I'm not sure it is easy, but the whole point is "just" to reduce the problem to an actual concave function — for which (2) might be easier))


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