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May 31, 2022 at 6:52 comment added J.Mayol Looks like a change of variable could be suffient: doing $x \mapsto x+1$ then we need to localize the new $x$ as $x = \mathcal{O}(\varepsilon)$, so introducing $x=\varepsilon \tilde{x}$ we obtain $(\varepsilon^{-3}h^2D_{\tilde{x}}^2 + \tilde x)u=0$ (changing charts, etc.). But for $\varepsilon= Ch^{2/3}$ we have $\varepsilon^{-3}h^2$ which is not a semiclassical parameter anymore ...
May 26, 2022 at 7:00 comment added J.Mayol I edited so that we only consider the $h_n$ values. What I want is the $O(\varepsilon^{1/4})$ bound that I'm asking, which is an upper bound (of course if you really look at the precise asymptotics of Hermite functions you will see that in this special case this is also a lower bound, but here I only want the upper bound.
May 26, 2022 at 2:29 comment added Willie Wong My (somewhat facetious) point is that there does exist a much simpler proof, albeit not the one you are looking for. So as the question asker, it may help if you specify more precisely what is the "lower bound" of simplicity you are looking for.
May 25, 2022 at 19:02 history edited J.Mayol CC BY-SA 4.0
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May 25, 2022 at 19:01 comment added J.Mayol @WillieWong the thing is to not use what is known about the Hermite polynomials. In fact, because it is semiclassical analysis one can change $x^2$ for some $V(x)$ with similar properties. For the fact that $Pu_h$ is not always solvable, just take $h=h_n=\frac{1}{\sqrt{2n+1}}$ (note that there is a square root missing in your $h$, it's because you should also rescale de $x^2$).
May 24, 2022 at 20:11 history edited J.Mayol CC BY-SA 4.0
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May 24, 2022 at 20:10 comment added J.Mayol ahh right, I'm assuming that, so I will edit and add it, thanks
May 24, 2022 at 18:30 history asked J.Mayol CC BY-SA 4.0