= Equality in Hölder's inequality
For $1<p,q<\infty$ and nonzero factors in their respective <Lp spaces>, equality in the product-integral bound holds precisely when
$$
\frac{|f|^p}{\lVert f\rVert_p^p}
=\frac{|g|^q}{\lVert g\rVert_q^q}
\quad\text{almost everywhere}.
$$
In other words, the normalized powers match. Zero factors give the trivial equality separately. For real <vectors>, equality for the signed finite <sum> also needs the nonzero coordinate products to have a common sign; otherwise the preceding <triangle inequality> loses information.
For $f(t)=t$ and $g(t)=1-t$ on $[0,1]$, with $p=3$, $q=3/2$,
$$
\int_0^1 f(t)g(t)\,dt=\frac16,
\qquad
\lVert f\rVert_3\lVert g\rVert_{3/2}
=\left(\frac14\right)^{1/3}\left(\frac25\right)^{2/3}
=25^{-1/3}\approx0.341995.
$$
Changing only $g$ to $t^2$ gives $f^3=g^{3/2}=t^3$, and both sides become $1/4$. The factors need not be equal: their conjugate powers are what must agree after normalization.
\Image[/holder-integral-products-and-equality.png]
{title=Matching conjugate powers makes the integral bound exact}
{description=Original <function graphs> on $[0,1]$. The green <area> is the <integral> of the product. For $g=1-t$, the normalized powers differ and the bound is strict. For $g=t^2$, $f^3=g^{3/2}$ pointwise, so the product <integral> attains the <Hölder's inequality> bound.}
The proof outline makes the equality condition natural. Normalize $u=|f|/\lVert f\rVert_p$ and $v=|g|/\lVert g\rVert_q$. The <Young inequality> gives $uv\leq u^p/p+v^q/q$; integration yields $\int uv\leq1/p+1/q=1$. Equality requires $u^p=v^q$ <almost everywhere>. The endpoint proof uses the <essential supremum> directly, so it does not need division by $p-1$ at $p=1$.
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