# Talk:Square-integrable function

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The article says, "The quadratically integrable functions form an inner product space." I'm pretty sure that isn't true! A function that differs from the zero function on a non-empty set of measure zero will have inner product zero, violating positive-definiteness. That's why you have to mod out by these functions when you define Lp spaces, right? ---Vectornaut (talk) 01:44, 29 May 2011 (UTC)

Two functions may attain different values on a set of measure zero and still be considered equal under any ${\displaystyle L_{p}}$ metric. Only the ${\displaystyle L_{2}}$ metric is induced by an inner product, because the polarization identity does not hold for other values of ${\displaystyle p}$. Therefore the other ${\displaystyle L_{p}}$ spaces are not inner product spaces. The equality of functions in the context of Lebesgue integration is redefined in terms of the Lebesgue integral of their difference. Therefore, even if they do differ on a set of Lebesgue measure zero, they are equal under the revised equality. This is how an ${\displaystyle L_{p}}$ space can be a metric space and how an ${\displaystyle L_{2}}$ space is an inner product space. So the condition of sign definiteness (or positive definiteness, although that is redundant and can be derived by sign definiteness and the triangle inequality) holds for the ${\displaystyle L_{p}}$ metrics under the revised equality. For more details see Naylor and Sell, Linear Operator Theory in Engineering and Science, Springer. — Preceding unsigned comment added by Helptry (talkcontribs) 03:21, 29 November 2012 (UTC)

@Helptry What you write is blatently wrong (very unusual at least). Two functions are equal, if their graphs are equal. What happens here is that the vector space N of Null-functions (functions whose support is a measure-0-set) must be (is) factored out from the space of sq-integrable functions cL2 (curlyL). Thus L2=cL2/N. The vector space of classes f+N, f sq. int. (f in cL2), then becomes a Hilbert space. This is correctly described in the article about L2 on WP. What you describe as "new equality" is equality in the quotient space cL2/N. — Preceding unsigned comment added by LMSchmitt (talkcontribs) 19:53, 24 November 2021 (UTC)

Reference in regard to the latter: https://en.wikipedia.org/wiki/Lp_space — Preceding unsigned comment added by LMSchmitt (talkcontribs) 20:00, 24 November 2021 (UTC)

## Why does L2 norm redirect here!?

I'm looking for an explanation of L2 norm. I get here instead and I see L2 and I see norm, but if you are going to redirect me here then I would like to see L2 norm. — Preceding unsigned comment added by 108.80.114.110 (talk) 11:42, 10 July 2014 (UTC)

For what it's worth, L2 norm redirects to Norm (mathematics)#Euclidean norm, and L2-norm redirects to Square-integrable function. I think L2-norm should be fixed to redirect to Norm (mathematics)#Euclidean norm, too. – Tobias Bergemann (talk) 11:57, 10 July 2014 (UTC)

There should be a redirect to https://en.wikipedia.org/wiki/Lp_space which is much better written then this article here. — Preceding unsigned comment added by LMSchmitt (talkcontribs) 20:03, 24 November 2021 (UTC)

## "L2 space" listed at Redirects for discussion

A discussion is taking place to address the redirect L2 space. The discussion will occur at Wikipedia:Redirects for discussion/Log/2020 June 10#L2 space until a consensus is reached, and readers of this page are welcome to contribute to the discussion. 1234qwer1234qwer4 (talk) 16:58, 10 June 2020 (UTC)

## Deleting some (correct) rubbish

In https://en.wikipedia.org/wiki/Lebesgue_integration#Lebesgue_integrable integrable is defined via finite Lebesque integral of the <|absolute|> value of a measerable function. Not using the positive and negative parts of Re(f) and Im(f). Thus the (mathematically correct, but too technical and trivial) section

"An equivalent definition is to say that the square of the function itself (rather than of its absolute value) is Lebesgue integrable. For this to be true, the integrals of the positive and negative portions of the real part must both be finite, as well as those for the imaginary part."

is unnecessary rubish. Replace by

Since ${\displaystyle |f(x)|^{2}=|f(x)^{2}|}$ , an equivalent definition is obviously to say that the square of the function is Lebesgue integrable. https://en.wikipedia.org/wiki/Lebesgue_integration#Lebesgue_integrable — Preceding unsigned comment added by LMSchmitt (talkcontribs) 20:41, 24 November 2021 (UTC)