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Lp Spaces and the Inequalities They Rest On

Functions with integrable p-th power form a normed space once functions that agree almost everywhere are treated as equal. Holder's and Minkowski's inequalities are what make the norm behave like a length.

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What a learner can do afterwards

  • Use Holder's inequality to bound the integral of a product
  • Explain the identification of functions agreeing almost everywhere and why the norm forces it
  • Say what makes the case p equal to 2 different from every other p

1 · Read

You measure a function by its Lp norm: integrate its p-th power and take the p-th root. Minkowski proved the triangle inequality for this measurement, which is why the space of functions with finite norm behaves like a geometry with lengths. Holder bounds the integral of a product by the product of the norms, using conjugate exponents whose reciprocals sum to 1.

Try it together

The indicator of the interval from 1 to 4 has L1 norm equal to its length, which is 3. The constant 3 on an interval of length 4 has L2 norm equal to the square root of 9 times 4, which is 6. You compute both directly from the definition of the norm.

You must treat two functions that agree except on a null set as the same element of Lp. The reason is definiteness: the indicator of a single point is nonzero somewhere yet its integral is 0, so without this step a nonzero function would carry norm 0. Identifying almost everywhere equal functions repairs the norm.

Good to know

When p equals 2, the norm comes from an inner product, so you gain orthogonality and the parallelogram law, and the space is a Hilbert space. For every other exponent no inner product sits behind the norm. As a quick check for Holder, the conjugate of 4 is 4 over 3, since 1 over 4 plus 3 over 4 equals 1.

Holder and Minkowski make Lp a normed geometry, almost everywhere equal functions count as one element, and only p equal to 2 brings an inner product.

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Lp Spaces and the Inequalities They Rest On · Mathematics, ages 22 to 24 · LightMySky