Let f be a strictly positive Schwartz function on $\mathbb R$. Does it imply $\sqrt f$ is a Schwartz function on $\mathbb R$?
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No. Differentiating $\sqrt{f(x)}$ gets degrees of $f$ in denominator and if $f$ approaches in some points to zero too fast it could lead to growth of derivatives. Consider function $f(x)=e^{-x^2} \left(e^{-x^2}+\sin ^2(x)\right)$. Using Mma for calculations we have $$ g(x)=\left(\sqrt{f(x)}\right)''= $$ $$ \frac{4 x^2+e^{2 x^2} \left(x^2-2\right) \sin ^4(x)+2 e^{x^2} \left(3 x^2-2\right) \sin ^2(x)+e^{x^2} \cos ^2(x)-2 e^{2 x^2} x \sin ^3(x)\cos(x)-2}{e^{x^2}\left(e^{x^2} \sin ^2(x)+1\right)^{3/2}} $$ and $$ g(k\pi)=e^{-(2 \pi k+\pi )^2} \left(4 (2 \pi k+\pi )^2+e^{(2 \pi k+\pi )^2}-2\right), $$ so $\lim_{k\to\infty}g(k\pi)=1\,$. The forth derivative of $\sqrt{f(x)}$ is unbounded: $$ g''(k\pi)= 4 e^{-\pi ^2 k^2} \left(4 \pi ^4 k^4-12 \pi ^2 k^2+3\right)-3 e^{\pi ^2 k^2}-4. $$
Andrew
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