My original question is if $f\in \mathcal{S}_{\alpha_1}^{\beta_1},\: g \in \mathcal{S}_{\alpha_2}^{\beta_2}$, where does $(f\cdot g) (x)=f(x)g(x)$ belong ? where $\mathcal{S}_\alpha^\beta$ is defined here.

To address this we proceed as follows:

$f\in \mathcal{S}_{\alpha_1}^{\beta_1},\: g \in \mathcal{S}_{\alpha_2}^{\beta_2}\implies$ $$\sup_x|f(x)|\exp(k_1|x|^{1/\alpha_1}) < \infty \\ \sup_x|g(x)|\exp(k_2|x|^{1/\alpha_2}) < \infty$$ Multiplying both the inequalities we have $$\sup_x|f(x)g(x)|\exp(k_1|x|^{1/\alpha_1}+k_2|x|^{1/\alpha_2}) < \infty -(1)$$ Now, $k_1|x|^{1/\alpha_1} \le k_1|x|^{\max\{1/\alpha_1,1/\alpha_2\}}$. Similarly $k_2|x|^{1/\alpha_2} \le k_2|x|^{\max\{1/\alpha_1,1/\alpha_2\}}$.

hence $$\sup_x|f(x)g(x)|\exp(k_1|x|^{1/\alpha_1}+k_2|x|^{1/\alpha_2})\le\sup_x|f(x)g(x)|\exp(k|x|^{\max\{1/\alpha_1,1/\alpha_2\}}\:)-(2)$$ where $k=k_1+k_2$. My question: How do I show that $$\sup_x|f(x)g(x)|\exp(k|x|^{\max\{1/\alpha_1,1/\alpha_2\}}\:) < \infty$$ given $(1),\: (2)$. But before that is it right what I have done so far ?. Any help will be welcome. Intuitively I feel that $(f \cdot g)(x)$ will belong to $\mathcal{S}_{\max\{1/\alpha_1,1/\alpha_2\}}^{\max\{1/\beta_1,1/\beta_2\}}$ $\:$and thats why I was proceeding in that direction. Thanks in advance.

  • $\begingroup$ The estimate $|x|^a \leq |x|^{\max\{a,b\}}$ is only valid for $|x|\geq 1$ in general. Furthermore, what you actually need would be the reverse inequality of (2). $\endgroup$ – PhoemueX Sep 22 '15 at 7:12
  • $\begingroup$ @PhoemueX, I have reframed the question, so that it becomes much clearer. Now can you help me ?. You can assume $|x| \ge 1$ $\endgroup$ – creative Sep 22 '15 at 8:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.