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02:41
Yes,nice! I should have written $<x,y_{1}-y_{2}> = 0 \forall x$ implying that $y_{1}-y_{2} = 0$
03:04
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1) Searching in main gave me - this1
I think since $l^{p},p\neq2$ is not an inner product spce it is also not a Hilbert space
as a Hilbert space is an inner product space with a norm so taking contrapositive we get if "it is not an inner product space then it is not a Hilbert space" , is this correct?
2) this is nice too! this1
In the first two answers i notice taking a function $f$ and constructing a second function $g$ by shifting the domain like $g(x) = f(1-x)$,why this works? any mathematical reasonging following this?
any other function will not work?>
 
11 hours later…
14:11
Do we have that $L^1(R) \subset L^2(R)$?
Maybe it's worth having a look at these posts:
I hope that helps @user1770201
I see you have asked also in the main chat room:
in Mathematics, 3 mins ago, by user1770201
Do we have that $L^1(R) \subset L^2(R)$?
Maybe you'll get some response there, too.
14:30
in Mathematics, 4 mins ago, by user1770201
@MartinSleziak I checked out that link but can't figure out how to use that to determine whether L^1(R) is a subset of L^2(R)
in Mathematics, 4 mins ago, by user1770201
But I think it must so, since 1/x is in L^2(R) but not in L^1(R)
How did you get that 1/x is in L^2?
Most likely neither of the inclusions is true, you can try to have a look here: List of counter-examples to $\mathcal{L}_p(\mathbb{R})$ inclusions.
 
1 hour later…
15:43
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In case of $X =C_{0,0}$
set of sequences which has finite number of terms non zero
then for $x,y \in X$,$<x,y> = \sum_{j=1}^{\infty} x(j) y(j)$ define a inner product space?
perhaps not asthere will be a problem in satisfying $<x,y> = \bar{<y,x>}$
when a sequence has complex entries
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So $<x,y> = \sum_{j=1}^{\infty} x(j) \bar{y(j)}$ define an inner product space

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