= Minkowski inequality
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{title2=$\|f+g\|_p\leq\|f\|_p+\|g\|_p$}
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For <measurable functions> $f,g$ with values in the <real numbers> or <complex numbers> on the same <measure space> $(X,\mathcal A,\mu)$, and $1\leq p\leq\infty$, membership in the same <Lp space> gives
$$
f,g\in L^p(\mu)\quad\Longrightarrow\quad f+g\in L^p(\mu),\qquad
\lVert f+g\rVert_p\leq\lVert f\rVert_p+\lVert g\rVert_p.
$$
Thus adding two <functions> cannot make their combined <Lp norm> exceed the sum of their separate <Lp norms>. The <measure> need not be finite or a <probability measure>. The hypotheses require finite <Lp norms>; $p=\infty$ uses the <essential supremum> of the <absolute value>, not an ordinary pointwise <supremum>. For <complex numbers>, the <absolute value> is the <complex modulus>. As usual, <Lp spaces> identify <functions> equal <almost everywhere>. <Measure, Integration & Real Analysis by Sheldon Axler>[Axler, Measure, Integration & Real Analysis, §7A, Definition 7.1, pp. 194–195, and Theorem 7.14, p. 199]{ref}
The <Minkowski inequality> makes the <Lp norm> a workable notion of size: it supplies the <triangle inequality> needed to turn $d_p(f,g)=\lVert f-g\rVert_p$ into a <metric> on <Lp space>. Applied to $(f-h)+(h-g)$, it says that travelling through an intermediate <function> $h$ cannot shorten the direct distance measured by this <metric>. Applied to two error <functions>, it gives a bound on the total error even when the errors reinforce one another; <Minkowski inequality for error aggregation> gives explicit examples.
The geometry is already visible in the <Euclidean plane>. Under <counting measure> on two coordinates, the closed <unit balls> for the <L1 norm>, <Euclidean norm>, and <L-infinity norm> are respectively $|x|+|y|\leq1$, $x^2+y^2\leq1$, and $\max(|x|,|y|)\leq1$. Every one is a <convex set>: if $u,v$ have <Lp norm> at most one, the <Minkowski inequality> and <absolute homogeneity of a norm> give $\lVert tu+(1-t)v\rVert_p\leq t+(1-t)=1$ for $0\leq t\leq1$. The <Euclidean norm> has no flat boundary segments, whereas the other two <norms> do. This is the distinction captured by a <strictly convex normed space> and developed in <equality in Minkowski inequality>.
\Image[/lp-unit-balls-triangle-geometry.png]
{title=Closed <unit balls> for the <L1 norm>, <Euclidean norm>, and <L-infinity norm>}
{description=The red point is the midpoint of the two marked unit <vectors>. It lies inside the disk, but on the flat edge of the diamond or square. All three <unit balls> are <convex sets>; only the disk gives a <strictly convex normed space> among these planar examples.}
A short <Hölder's inequality> proof outline handles $1<p<\infty$. First $|f+g|^p\leq2^{p-1}(|f|^p+|g|^p)$, so $f+g$ has finite <Lp norm> before any division. With <conjugate exponents> $p$ and $q=p/(p-1)$, set $h=|f+g|^{p-1}$; then $\lVert h\rVert_q=\lVert f+g\rVert_p^{p-1}$. The scalar <triangle inequality> followed by <Hölder's inequality> gives
$$
\lVert f+g\rVert_p^p
\leq\int (|f|+|g|)|f+g|^{p-1}\,d\mu
\leq(\lVert f\rVert_p+\lVert g\rVert_p)\lVert f+g\rVert_p^{p-1}.
$$
Divide if $\lVert f+g\rVert_p>0$; otherwise the desired bound is immediate. For $p=1$, integrate $|f+g|\leq|f|+|g|$. For $p=\infty$, the <almost everywhere> bounds $|f|\leq\lVert f\rVert_\infty$ and $|g|\leq\lVert g\rVert_\infty$ give the result by the <essential supremum>. The same scalar <triangle inequality> works for <real numbers> and <complex numbers>.
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