Problem of the Week #245 - Mar 07, 2017

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Euge

MHB Global Moderator
Staff member
Here is this week's POTW:

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Consider the Lebesgue space $L^1(\Bbb R)$ as an algebra with product given by convolution. Prove that $L^1(\Bbb R)$ is isomorphic as an algebra to an ideal in the algebra $M(\Bbb R)$ of complex Borel measures on $\Bbb R$, and identify the ideal. Note the product in $M(\Bbb R)$ is given by convolution of measures.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

Euge

MHB Global Moderator
Staff member
I'm going to give members one extra week to solve this POTW. For a hint, consider the set of measures absolutely continuous with respect to the Lebesgue measure on $\Bbb R$. Use the Radon-Nikodym theorem to construct an algebra map.

Euge

MHB Global Moderator
Staff member
Honorable mention goes to vidyarth for an incomplete solution. You can read my solution below.

Let $A_m$ denote the set of measures absolutely continuous with respect the Lebesgue measure $m$. Then $A_m$ is an ideal of $M(\Bbb R)$. For given $\mu\in M(\Bbb R)$ and $\nu \in A$, $d\mu = f\, dm$ for some $f\in L^1(\Bbb R)$, and for every measurable set $E\subset \Bbb R$,

$$(\mu * \nu)(E) = \int \mu(E - t)\, d\nu(t) = \iint 1_{E-t}(s)f(s)\, dm(s)\, d\nu(t) = \iint 1_{E}(t + s)f(s)\, d\nu(t)\, dm(s)$$
$$= \iint 1_E(s) f(s-t)\, d\nu(t)\, dm(s) = \int_E (f * \nu)(s)\, dm(s)$$

where $(f*\nu)(s) = \int f(s-t)\, d\nu(t)$. Thus $d(\mu * \nu) = (f * \nu)\, dm$, proving $\mu * \nu \in A_m$.

By the Radon-Nikodym theorem, the mapping $\Phi : A_m \to L^1(\Bbb R)$, $\mu \mapsto \frac{d\mu}{dm}$ is a one-to-one correspondence. For all $\mu, \nu \in A_m$,

$$\frac{d(\mu * \nu)}{dm} = \frac{d\mu}{dm} * \frac{d\nu}{dm}$$

whence $\Phi$ is an isomorphism of algebras.

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