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If I have a function $f$ belongs to the Schwartz space, i.e. $f\in \mathcal{S}$, how can I prove $f\in L^p$ ?

I know that $\mathcal{S}\subset L^p$ hence the above should make sense. But I need a proof.

If I just take a function from $\mathcal{S}$, then compute its $L^p$ norm, and I can see clearly it is bounded hence $f\in L^p$, I mean $$ \int|e^{-a|x|^2}|^p<\infty$$ is obvious. But is this a proof?

math101
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A function in $\mathcal{S}$ decays faster than the reciprocal of any polynomial at infinity and is bounded. (This is usually built into the definition.) So its magnitude is bounded by some $M_1$ on the ball of some radius $R$ centered at the origin, and by $M_2/|x|^N$ outside this ball. Choose $N$ sufficiently large (depending on the dimension and $p$) to get your result.

Ian
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  • thanks. But I'm sort of looking for somewhat different answer. Trying to relate the proof of S(R) dense in L^p... – math101 Nov 11 '15 at 23:51
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    @math101 I'm not sure how an answer could be fundamentally different; to prove something is in $L^p$ you pretty much need to compare it to something else that you already know is in $L^p$. The density proof is quite different. – Ian Nov 11 '15 at 23:55
  • It would be good if you could write in term of the mathematics. I understand what you mean but my difficulty is how to write the mathematics precisely. (Because my applied background...) – math101 Nov 12 '15 at 05:15
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    @math101 What do you mean, exactly? Do you mean to write the proof formally? Well, you need $1/|x|^N$ (away from the origin, zero near it) to be in $L^p(\mathbb{R}^d)$. By writing in polar coordinates, it is equivalent to have $r^{d-1-N}$ in $L^p(\mathbb{R})$, so it is equivalent to have $r^{p(d-1-N)}$ in $L^1(\mathbb{R})$. But we know the cutoff there, we need the exponent to be less than $-1$. So $p(d-1-N)<-1$, hence $-N<\frac{-1}{p}-d+1$, so $N>\frac{1}{p}+d-1$. For instance with $p=2,d=3$ you need $N>\frac{5}{2}$. – Ian Nov 12 '15 at 05:45
  • sorry Ian, your answer is OK but it is quite different from what I expect. I thought the proof of this is the same as the proof of the fact $S(R)\subset L^p$ – math101 Nov 12 '15 at 06:04
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    @math101 Not at all. The density requires much more sophisticated machinery. – Ian Nov 12 '15 at 15:06