Prove the triangle inequality in Lp

Try

On a two-point space, adding f=(1,0) and g=(0,1) yields (1,1). Its Euclidean length sqrt (2) is at most the sum of lengths two. But if one tries the expression corresponding to p=1/2, the sum has size four while each input has size one. The condition p≥1 determines whether this family of expressions behaves like distance.

Predict

Predict whether a function with finite square integral must have finite absolute integral. On a probability space it does, by Hölder with the constant one. On the positive half-line it need not: 1/x on [1, infinity) has finite square integral but divergent absolute integral. The measure of the domain changes which spaces contain which others.

Experiment

For f=(1,0) and g=(0,1), compare the same pair under different exponents.

Exponent	Size of f	Size of g	Size of f+g
1	1	1	2
2	1	1	sqrt (2)
infinity	1	1	1
1/2	1	1	4

Explain why the last row is a useful counterexample rather than a new norm. Then rescale the counting measure to assign probability one half to each point and recalculate the first three rows.

Understand

Minkowski’s inequality states norm (f+g)_p≤norm (f)_p+norm (g)_p for 1≤p≤infinity. For p=1 it follows by integrating the pointwise triangle inequality. For infinity, combine almost-everywhere bounds outside the union of two null exceptional sets.

For 1< p< infinity, first establish that f+g belongs to Lp. Convexity of t^p gives abs (f+g)^p≤2^(p−1)(abs (f)^p+abs (g)^p), whose integral is finite. Now write abs (f+g)^p≤(abs (f)+abs (g))abs (f+g)^(p−1). Apply Hölder separately to the two terms, using exponents p and q=p/(p−1). The q norm of abs (f+g)^(p−1) is norm (f+g)_p^(p−1). If the sum norm is positive, divide by this finite quantity; if it is zero the result already holds. This proves the triangle inequality without assuming the desired finiteness in advance.

Consequently d(f, g)=norm (f−g)_p is a metric on almost-everywhere equivalence classes. Translation and scaling behave as expected. The quotient is essential: two representatives differing on a null set have zero distance.

On a finite-measure space, Hölder also compares different exponents. For p< q, apply it to abs (f)^p and 1 with exponents q/p and q/(q−p). Taking pth roots gives norm (f)_p≤μ(X)^(1/p−1/q)norm (f)_q. For probability measure the factor is one. For an infinite measure it is not a useful finite bound, and actual containments can fail. The case q=infinity follows directly from an essential bound times the finite domain measure.

Apply

Use the metric to interpret approximation error: the distance between two estimates is bounded by the sum of their distances to a reference. This is useful when splitting an approximation into truncation error, simple-function error and a continuous approximation error. Each step must be measured in the same norm; small pointwise error on an infinite domain need not imply small Lp error.

Continue

The triangle inequality lets us control sums of errors, but it does not yet prove that every Cauchy sequence has a limit in the space. That completeness statement is next. Its proof constructs a rapidly convergent subsequence and turns a summable sequence of norm errors into almost-everywhere convergence.

Continue to Construct limits of Cauchy sequences.