= Finite-vector example of Hölder's inequality
Take $p=3$, $q=3/2$, and <vectors> $x=(1,2,-1)$ and $y=(2,-1,3)$. Their coordinate products are $(2,-2,-3)$, so
$$
\left|\sum_i x_i y_i\right|=3,\qquad
\sum_i|x_i y_i|=7.
$$
Their <Lp norms> give the larger bound
$$
\lVert x\rVert_3\lVert y\rVert_{3/2}
=10^{1/3}(2\sqrt2+1+3\sqrt3)^{2/3}
\approx9.338662.
$$
Thus cancellation explains $3<7$, and <Hölder's inequality> explains the remaining gap between $7$ and the norm product. Small signed <sums> need not imply small individual coordinate products.
Now keep $x$ and choose $y=(1,4,-1)$. The coordinate products are $(1,8,1)$, with no cancellation, and
$$
\sum_i x_i y_i=10
=10^{1/3}10^{2/3}
=\lVert x\rVert_3\lVert y\rVert_{3/2}.
$$
The matching powers are $|x_i|^3=|y_i|^{3/2}$, namely $(1,8,1)$.
\Image[/holder-vector-products-and-norm-bounds.png]
{title=Cancellation and Hölder's inequality account for two different gaps}
{description=Two original finite-vector calculations with $p=3$, $q=3/2$. The first pair has signed <sum> $-3$, absolute-product <sum> $7$ and norm-product bound approximately $9.339$. The second pair attains equality at $10$. The diagrams show calculated values, with no simulated data.}
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