Lipschitz-type estimate... True or false?

Question

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!

Answer 1

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 $\alpha$.

Answer 2

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.