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 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 space.

Specific counterexamples

To show that the parallelogram law

holds iff , the following counterexamples may be used

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