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Prove that the set of all continuous functions $ R = \{f: \mathbb{R} \rightarrow \mathbb{R} \} $ on $ \mathbb{R} $ satisfying

$ \int_{ \mathbb{R} } | f(x) | dx < \infty$

forms a ring.

It's easy to see these functions form an abelian group with respect to the usual addition of functions. However in this case we can't use the usual multiplication operation $(f \times g) (x) = f(x)g(x)$ since our identity would have to be $1$ which doesnt have a finite integral over the whole of $\mathbb{R}$ so is not in $R$.

I tried to pick a random function, say $ 1/ (1+x^2) $ and 'force' it to be the identity somehow but I couldn't find a way.

Would I possibly have to come up with another addition operation as well?

Any hints would be appreciated.

Evgeny T
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  • Why must all rings have an identity? The even integers have no identity but are a ring. – CPM Dec 17 '16 at 13:58
  • @CPM They have a multiplicative and additive identity no? In integers we have 1 and 0 respectively. – Evgeny T Dec 17 '16 at 14:03
  • Not all rings have to have a multiplicative identity. Usually you just need multiplication to be associative and distribute over addition. May well not be any identity and thus inverses of elements don't even make sense. – CPM Dec 17 '16 at 14:06
  • There are some people who insist that rings do have a multiplicative identity as a ring axiom, which is your definition of ring? $2\mathbb{Z}$ has no identity, but in my mind is a pretty standard ring example – CPM Dec 17 '16 at 14:09
  • @CPM Ah yes ok. Well since the question is easy in the case we don't need an identity let's say there must be one. – Evgeny T Dec 17 '16 at 14:10
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    There can be some known definitions of such operations, but I can offer the following one: Let $(f + g)(x)$ equal $f(x) + g(x)$, and $(f\cdot g)(x)$ equal $\int_\mathbb{R}f(t)g(x - t),dt$. Thus if you want an identity, you have to take a generalized $\delta$-function. – Golovanov399 Dec 17 '16 at 14:55
  • @Golovanov399 Ah yes I did have a look at using the delta function but didnt define the product in exactly this form. Thanks. – Evgeny T Dec 17 '16 at 15:21
  • It is interesting to note that any ring without identity (rng, a ring without the i) can be embedded into a ring with identity using a Dorrah Extension. This post has some information about it that you might find interesting, especially in that top answer. Some really great stuff there: http://math.stackexchange.com/questions/16168/applications-of-rings-without-identity – CPM Dec 17 '16 at 17:07

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