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Given that two formal power series have coefficients in $\ell^p$,

$$a(x)=\sum_{n=0}^\infty a_nx^n,\quad\sum_{n=0}^\infty|a_n|^p<\infty$$

$$b(x)=\sum_{n=0}^\infty b_nx^n,\quad\sum_{n=0}^\infty|b_n|^p<\infty,$$

does it follow that their product $a(x)b(x)$ also has coefficients in $\ell^p$?

I'm particularly interested in the case $p=2$ (Hilbert space).

I've found that the multiplication operator is unbounded with respect to the $\ell^2$ norm; consider the polynomials

$$p_n(x)=\frac{1-x^n}{1-x}=1+x+x^2+\cdots+x^{n-1}.$$

$$\lVert p_n\rVert_2^2=n$$

$$p_n(x)^2=1+2x+3x^2+\cdots+(n-1)x^{n-2}+nx^{n-1}+(n-1)x^n+\cdots+2x^{2n-3}+x^{2n-2}$$

$$\lVert p_n\!^2\rVert_2^2=2(1^2+2^2+3^2+\cdots+n^2)-n^2$$

$$=2\frac{n(n+1)(2n+1)}{6}-\frac{n(3n)}{3}$$

$$=\frac{n(2n^2+1)}{3}$$

So for all $n$,

$$\sup_{a,b}\frac{\lVert ab\rVert_2}{\lVert a\rVert_2\lVert b\rVert_2}\geq\sup_a\frac{\lVert a^2\rVert_2}{\lVert a\rVert_2^2}\geq\frac{\lVert p_n\!^2\rVert_2}{\lVert p_n\rVert_2^2}=\sqrt\frac{2n^2+1}{3n}$$

and thus $$\sup_{a,b}\frac{\lVert ab\rVert_2}{\lVert a\rVert_2\lVert b\rVert_2}=\infty.$$

But is this supremum actually a maximum? Is there any case where $\lVert a\rVert_2$ and $\lVert b\rVert_2$ are finite, but $\lVert ab\rVert_2$ is infinite?

mr_e_man
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1 Answers1

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For $p\leq1$, using the concavity condition $(|x|+|y|)^p\leq|x|^p+|y|^p$, we have

$$\lVert ab\rVert_p^p=|a_0b_0|^p+|a_1b_0+a_0b_1|^p+|a_2b_0+a_1b_1+a_0b_2|^p+\cdots$$

$$\leq|a_0b_0|^p+(|a_1b_0|+|a_0b_1|)^p+(|a_2b_0|+|a_1b_1|+|a_0b_2|)^p+\cdots$$

$$\leq|a_0b_0|^p+|a_1b_0|^p+|a_0b_1|^p+|a_2b_0|^p+|a_1b_1|^p+|a_0b_2|^p+\cdots$$

$$=(|a_0|^p+|a_1|^p+|a_2|^p+\cdots)(|b_0|^p+|b_1|^p+|b_2|^p+\cdots)$$

$$=\lVert a\rVert_p^p\lVert b\rVert_p^p$$

$$<\infty$$

so $\ell^p$ is closed under multiplication. (And $\ell^1$ is also a Banach algebra.)

For $p>1$, consider for example the function

$$a_3(x)=\frac{1}{\sqrt[3]{1-x}}$$

whose third power's coefficients are clearly not in any $\ell^p$ (except $\ell^\infty$, but even that is lost by squaring again) :

$$a_3(x)^3=\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$

But $a_3$ itself is in some $\ell^p$, which would therefore not be closed under multiplication. Indeed, from the binomial theorem,

$$a_3(x)=1+\Big(\frac13\Big)x+\Big(\frac13\cdot\frac43\Big)\frac{x^2}{1\cdot2}+\Big(\frac13\cdot\frac43\cdot\frac73\Big)\frac{x^3}{1\cdot2\cdot3}+\cdots$$

$$=1+\frac13x+\frac{1\cdot4}{3\cdot6}x^2+\frac{1\cdot4\cdot7}{3\cdot6\cdot9}x^3+\cdots.$$

Now, taking inspiration from this question to use the AM-GM inequality,

$$3n=\frac{(3n-2)+(3n+1)+(3n+1)}{3}\geq\sqrt[3]{(3n-2)(3n+1)(3n+1)}$$

so the coefficients of $a_3(x)$ are

$$\frac{1\cdot4\cdot7\cdots(3n-5)\cdot(3n-2)}{3\cdot6\cdot9\cdots(3n-3)\cdot(3n)}$$

$$\leq\frac{1\cdot4\cdot7\cdots(3n-5)\cdot(3n-2)}{(\sqrt[3]{1}\cdot\sqrt[3]{4}^2)\cdot(\sqrt[3]{4}\cdot\sqrt[3]{7}^2)\cdots(\sqrt[3]{3n-5}\cdot\sqrt[3]{3n-2}^2)\cdot(\sqrt[3]{3n-2}\cdot\sqrt[3]{3n+1}^2)}$$

$$=\frac{1}{\sqrt[3]{3n+1}^2}$$

which implies

$$\lVert a_3\rVert_p^p\leq\sum_{n=0}^\infty\frac{1}{(3n+1)^{2p/3}}.$$

This is finite for $\frac23p>1$, that is, $p>\frac32$. So for any such $p$ (including $p=2$), $a_3$ is in $\ell^p$ and $a_3\!^3$ is not.

Similar arguments applied to the functions

$$a_m(x)=\frac{1}{\sqrt[m]{1-x}}$$

show that, for $p>\frac{m}{m-1}$, $a_m$ is in $\ell^p$ and $a_m\!^m$ is not. Therefore, for any $p>1$, $\ell^p$ is not closed under multiplication.

mr_e_man
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