L1 is not an inner product space?
sto
I can show that S is an inner product space iff the norm ||.|| satisfies
the (parallelogram) equality
||f + g||^2 + ||f-g||^2 = 2 ( ||f||^2 + ||g||^2 )
in which case I can then prove that L1([0,2pi],m), where m is Lebesque
measure on the real line, is not an inner product space by using the
counter example
f(x) = 1
g(x) = sin(x)
and showing that the integrals in the RHS and LHS of the parallelogram
equality are not equal.
Is there a way to generalize this argument so that it does not deal
specifically with L1([0,2pi]), but rather is a general statement about
L1(X,m) where (X,m) is an arbitrary measure space?
Thanks,
-sto
José Carlos Santos
Suppose that there are two disjoint measurable subsets A and B of X
whose measure is finite and greater than 0. Let _f_ be the
characteristic function of A and let _g_ be the characteristic function
of B. Then
||f + g||^2 + ||f - g||^2 = 2(m(A) + m(B))^2
and
2(||f||^2 + ||g||^2) = 2(m(A)^2 + m(B)^2).
The second value is smaller than the first one.
Best regards,
Jose Carlos Santos
David C. Ullrich
Read what Arturo said. Then also note that there do exist measure
spaces for which L^1 _is_ an inner-product space... (not very
interesting measure spaces, but their existence says something
about what you're going to be able to prove in this direction).
>Thanks,
>-sto
José Carlos Santos
>> Let S be a normed space with norm ||.||.
>>
>> I can show that S is an inner product space iff the norm ||.|| satisfies
>> the (parallelogram) equality
>>
>> ||f + g||^2 + ||f-g||^2 = 2 ( ||f||^2 + ||g||^2 )
>>
>> in which case I can then prove that L1([0,2pi],m), where m is Lebesque
>> measure on the real line, is not an inner product space by using the
>> counter example
>>
>> f(x) = 1
>> g(x) = sin(x)
>>
>> and showing that the integrals in the RHS and LHS of the parallelogram
>> equality are not equal.
>>
>> Is there a way to generalize this argument so that it does not deal
>> specifically with L1([0,2pi]), but rather is a general statement about
>> L1(X,m) where (X,m) is an arbitrary measure space?
>
> Read what Arturo said.
Are you confusing Arturo with me? Well, I guess that's a compliment... :-)
David C. Ullrich
Sorry, you sci.math guys all look alike.
sto
Denote by R the real line, by _S_ the sigma algebra {0,R}, and by m
Lebesque measure on _S_. Then the only functions measurable _S_ are the
constant functions f(x) = c, c in R, and the only one of these in
L1(X,m) is f(x) = 0. So given any f,g in L1, f=g=0 in which case
|| f+g ||^2 + || f-g ||^2 = 0 = 2 ( || f ||^2 + || g ||^2 )
which implies L1(X,m) is an inner product space.
But is it the *only* one (with an L1 norm)?
sto
That makes sense, thanks.
So a sufficient condition for L1(X,m) *not* to be an inner product space
is that there exist two disjoint measurable subsets of finite, nonzero
measure. Can we go further and say that this is a necessary condition?
It seems plausible to me. In my reply to David Ullrich I give the
example of a (trivial) L1 space that is also an inner product space, but
it clearly does not have two disjoint subsets of finite, non-zero measure.
José Carlos Santos
>>> Let S be a normed space with norm ||.||.
>>>
>>> I can show that S is an inner product space iff the norm ||.|| satisfies
>>> the (parallelogram) equality
>>>
>>> ||f + g||^2 + ||f-g||^2 = 2 ( ||f||^2 + ||g||^2 )
>>>
>>> in which case I can then prove that L1([0,2pi],m), where m is Lebesque
>>> measure on the real line, is not an inner product space by using the
>>> counter example
>>>
>>> f(x) = 1
>>> g(x) = sin(x)
>>>
>>> and showing that the integrals in the RHS and LHS of the parallelogram
>>> equality are not equal.
>>>
>>> Is there a way to generalize this argument so that it does not deal
>>> specifically with L1([0,2pi]), but rather is a general statement about
>>> L1(X,m) where (X,m) is an arbitrary measure space?
>>
>> Suppose that there are two disjoint measurable subsets A and B of X
>> whose measure is finite and greater than 0. Let _f_ be the
>> characteristic function of A and let _g_ be the characteristic function
>> of B. Then
>>
>> ||f + g||^2 + ||f - g||^2 = 2(m(A) + m(B))^2
>>
>> and
>>
>> 2(||f||^2 + ||g||^2) = 2(m(A)^2 + m(B)^2).
>>
>> The second value is smaller than the first one.
>
> That makes sense, thanks.
> So a sufficient condition for L1(X,m) *not* to be an inner product space
> is that there exist two disjoint measurable subsets of finite, nonzero
> measure. Can we go further and say that this is a necessary condition?
> It seems plausible to me. In my reply to David Ullrich I give the
> example of a (trivial) L1 space that is also an inner product space, but
> it clearly does not have two disjoint subsets of finite, non-zero measure.
Yes, it is a necessary condition.
Suppose that it is _not_ true that there are two disjoint measurable
subsets whose measure is finite and greater than 0. Of course, this
will happen when the measure of any measurable set is either 0 or +oo,
and then L^1 will be an inner product space.
Otherwise, there is a measurable set A whose measure is finite and
greater than 0. But then, for any measurable set B whose measure is
finite and greater than 0, A and B have non-empty intersection. It
follows from this that a function _f_ is in L^1 if and only if the
restriction of _f_ to A is almost constant (that is, it is constant
outside a set with measure 0) and _f_ is almost always 0 in X\A.
Then||f||_1 = |k|, where _k_ is the constant such that f|_A = k almost
always. This space is then an inner product space.
An example of such a space is X = {0,1,2} and _m_ such that m({0}) = 0,
m({1}) = 1, and m({2}) = +oo.
David C. Ullrich
Ok, that's an example where L^1 is a zero-dimensional
inner product space. Hint: It's easy to give examples
where L^1 is a one-diimensional inner product space.
I haven't thought about it in detail, but I'm pretty
sure that more or less what Carlos said shows that
if the dimension of L^1(mu) is greater than 1 then
it is not an inner-product space.
sto
I can see that if f is constant for almost all x in A (and a.e. zero
outside), then f is in L1, because || f ||_1 = |k| m(A) and A has finite
measure. But I don't see how to prove the reverse implication: f in L1
then => f is constant a.e. in A. I can see that if f is measurable and
*not* constant a.e. on A, and in addition f is restricted to taking on
only two distinct values c1 and c2 on A, then the sets {x:f(x) = c1} /\
A and {x:f(x) = c2} /\ A are two disjoint measurable sets of nonzero
finite measure, which contradicts the assumption no such sets exist.
But without the restriction that f takes on only finitely many values on
A, I don't see how to construct a proof of the reverse implication.
> Then||f||_1 = |k|, where _k_ is the constant such that f|_A = k almost
> always.
Do you mean || f ||_1 = |k| m(A) ?
sto
It seems to me that José's example is a one dimensional space; it
contains the zero function and is spanned by the function f(x) =
I_{2}(x), where I_{2} is the characteristic function of the set {2}.
José Carlos Santos
Consider the map _g_ from R into [0,m(A)] defined by
g(x) = m(f^{-1}((-oo,x))).
Of course, _g_ is an increasing function. Saying that _f_ is almost
always equal to some number _k_ is the same thing as saying that g(x)
is 0 when x <= k and that g(x) = m(A) when x > k.
Now, suppose that _f_ isn't almost constant. Then there will be a
number _k_ such that g(k) is neither 0 nor m(A). Consider the set
B = f^{-1}((-oo,k)). Then m(B) = g(k), which is neither 0 nor +oo and
m(A\B) = m(A) - g(k), which is also neither 0 nor +oo. Therefore,
B and A\B have non-empty intersection, which is absurd.
>> Then||f||_1 = |k|, where _k_ is the constant such that f|_A = k almost
>> always.
>
> Do you mean || f ||_1 = |k| m(A) ?
Yes.
> This space is then an inner product space.
Best regards,
Jose Carlos Santos
David C. Ullrich
Yes it is - I hadn't looked at that post until now.