I ##L^2## - space equivalence classes and norm

[cianfa72]

TL;DR Summary

$L^2$-space as set of equivalence classes of squared integrable measurable functions

$L^2$-space is defined as equivalence classes on the set $\mathcal L^2$ of squared integrable measurable functions $f$ defined on the measure space $(\Omega, \mathcal A, \mu)$.

The equivalence relation $\sim$ is: $f \sim g$ iff $f=g$ almost everywhere (a.e.).

Prove that the above is equivalent to $||f|| = ||g||$. The norm $|| \, . ||$ is defined as $$\int_{\Omega} {| \, . |}^2 \, d{\mu}$$
My attempt
Consider the map $|f| - |g|$ that is measurable since $f,g$ are. $|f| - |g| \neq 0$ on a subset of the (measurable) set $A$ where $(f - g)$ is, therefore such subset (that is measurable itself) has measure 0. Hence $$\int_{\Omega} {|f| - |g|} \, d{\mu} = 0$$ i.e. by linearity $$\int_{\Omega} {|f|} \, d{\mu} = \int_{\Omega} {|g|} \, d{\mu}$$
Same argument applies to $|f|^2 - |g|^2$.

Does the above make sense ? Thanks.

 

[pasmith]

TL;DR Summary: $L^2$-space as set of equivalence classes of squared integrable measurable functions

$L^2$-space is defined as equivalence classes on the set $\mathcal L^2$ of squared integrable measurable functions $f$ defined on the measure space $(\Omega, \mathcal A, \mu)$.

The equivalence relation $\sim$ is: $f \sim g$ iff $f=g$ almost everywhere (a.e.).

Prove that the above is equivalent to $||f|| = ||g||$. The norm $|| \, . ||$ is defined as $$\int_{\Omega} {| \, . |}^2 \, d{\mu}$$

This is not a norm: you need to take the square root of it for it to satisfy the property \|af\| = |a|\|f\| for constant a.

I think the equivalence that you are looking for is that f = g a.e. is equivalent to \|f - g\| = 0.

f = g almost everywhere implies \|f\| = \|g\|, but the converse does not hold: Take f = 1, g = -1 and \Omega = [0,1] for a counterexample.

One of the axioms of a norm is that \|f\| = 0 if and only if f is the zero vector; for functions under pointwise addition the zero vector is the function which is identically zero. But for the L^2 norm, a function which is different from zero on a set of measure zero - and is therefore distinct from the zero vector - will have zero norm. We fix this by working instead with equivalance classes, with functions defined to be equivalent if and only if they differ from each other by a function with zero norm, i.e. f \sim g \quad\Leftrightarrow \quad \|f - g\| = 0.

 

[cianfa72]

This is not a norm: you need to take the square root of it for it to satisfy the property \|af\| = |a|\|f\| for constant a.

Ok yes, definitely. It is actually $$\left( \int_{\Omega} {| \, . |}^2 \, d{\mu} \right)^{\frac 1 2}$$

f = g almost everywhere implies \|f\| = \|g\|, but the converse does not hold: Take f = 1, g = -1 and \Omega = [0,1] for a counterexample.

Ok, yes.

One of the axioms of a norm is that \|f\| = 0 if and only if f is the zero vector; for functions under pointwise addition the zero vector is the function which is identically zero. But for the L^2 norm, a function which is different from zero on a set of measure zero - and is therefore distinct from the zero vector - will have zero norm. We fix this by working instead with equivalance classes, with functions defined to be equivalent if and only if they differ from each other by a function with zero norm, i.e. f \sim g \quad\Leftrightarrow \quad \|f - g\| = 0.

Ok, in the specific case the latter equivalence boils down to ${|f - g|}^2 \neq 0$ if and only if $f \neq g$. Therefore both hold true on the same (measurable) set $A$ with measure 0. This is true for any p-norm.

 

[cianfa72]

We fix this by working instead with equivalance classes, with functions defined to be equivalent if and only if they differ from each other by a function with zero norm, i.e. f \sim g \quad\Leftrightarrow \quad \|f - g\| = 0.

BTW, I'd say they are defined to be equivalent iff they differ each other by a function with zero "semi-norm" i.e. w.r.t. the map that will be promoted to a full norm when considering the set of equivalence classes being defined.

 

[pasmith]

BTW, I'd say they are defined to be equivalent iff they differ each other by a function with zero "semi-norm" i.e. w.r.t. the map that will be promoted to a full norm when considering the set of equivalence classes being defined.

Technically yes, although in general usage one refers to \|\cdot\|_p : \mathcal{L}^p \to [0, \infty) : f \mapsto \left(\int_\Omega |f|^p\,d\mu\right)^{1/p} as a norm and writes \|f\|_p for what is technically \|[f]\|_p. It's also possible to find that both the set of functions for which \int_{\Omega} |f|^p\,d\mu exists and the set of equivalence classes of such functions are denoted by \mathcal{L}^p(\Omega).