Lebesgue space

forms an inner product space iff

Let be a measure space with at least two distinct subsets of finite nonzero measure and let . Then the Lebesgue space satisfies the parallelogram law and therefore has a unique inner product iff .

Proof

Without loss of generality we may assume that . Let and , and

Then using the indicator functions and we have

and

so for equality we require . If , then is a strictly convex function, since its second derivative is positive on its domain, so since [[Convexity on the positive reals and negative f(0) implies superadditivity|strict convexity on the positive reals and implies superadditivity]] we have

so equality cannot hold. For an analogous argument can be made using .

For the converse, see L2 space.

Specific counterexamples

To show that the parallelogram law

holds iff , the following counterexamples may be used

• For take and where .
• For , i.e. Lebesgue sequence space, take and

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