What Equality in Hölder’s Inequality Requires
Hölder’s inequality bounds the integral of a product by the product of two norms. Equality is more restrictive: the functions’ magnitudes must be matched in the precise proportions dictated by the conjugate exponents. For complex-valued functions, there is a second issue. Even if the magnitudes attain the bound, varying phases can make the integral of the product smaller through cancellation.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; it need not be finite, sigma-finite, or complete. Let \(1<p<\infty\), and let \(q=p/(p-1)\), so \(1/p+1/q=1\). The functions \(f:X\to\mathbb{C}\) and \(g:X\to\mathbb{C}\) are measurable, with \(f\in L^p(\mu)\) and \(g\in L^q(\mu)\). Real-valued functions are included as a special case. All equality conditions between functions below hold \(\mu\)-a.e.
The norms are \(\|f\|_p=(\int |f|^p\,d\mu)^{1/p}\) and \(\|g\|_q=(\int |g|^q\,d\mu)^{1/q}\). Hölder’s inequality gives \(\int |fg|\,d\mu\le\|f\|_p\|g\|_q\), so \(f\overline{g}\) is integrable and its integral is defined. The equality criterion for the nonnegative integral \(\int |fg|\,d\mu\) is as follows.
The normalized formulation is useful because both sides of the condition integrate to \(1\). The proportionality formulation makes the underlying shape condition more visible: the two powers must agree up to one constant, not merely be supported on the same set or have the same integral.
Why the Criterion Holds
The proof reduces equality in the integral inequality to equality in Young’s inequality at almost every point. For nonnegative numbers \(a,b\), Young’s inequality says \(ab\le a^p/p+b^q/q\). Its equality condition is \(a^p=b^q\). To see the equality condition, fix \(a\ge0\) and consider \(H(b)=b^q/q-ab+a^p/p\) for \(b\ge0\). If \(a>0\), its minimum occurs where \(H'(b)=b^{q-1}-a=0\), namely \(b=a^{p-1}\); at that value \(b^q=a^p\) and \(H(b)=0\). If \(a=0\), \(H(b)=b^q/q\), which is zero exactly when \(b=0\). Thus the gap is nonnegative and vanishes exactly when \(a^p=b^q\).
Proof. First suppose both norms are positive, and set \(A=|f|/\|f\|_p\) and \(B=|g|/\|g\|_q\). These are nonnegative measurable functions with \(\int A^p\,d\mu=1\) and \(\int B^q\,d\mu=1\). By Hölder’s inequality, \(AB\) is integrable. Young’s inequality gives the nonnegative measurable function $$ D=\frac{A^p}{p}+\frac{B^q}{q}-AB\ge0. $$ It is integrable because each of its three terms is integrable. Its integral is $$ \int D\,d\mu=\frac1p+\frac1q-\int AB\,d\mu=1-\int AB\,d\mu. $$ Since \(\int AB\,d\mu=\int |fg|\,d\mu/(\|f\|_p\|g\|_q)\), equality in Hölder holds exactly when \(\int D\,d\mu=0\).
For any nonnegative measurable integrable function \(D\), \(\int D\,d\mu=0\) if and only if \(D=0\) \(\mu\)-a.e. Indeed, if the integral is zero, then for each positive integer \(n\), monotonicity of the integral gives \(\mu(\{D>1/n\})/n\le\int D\,d\mu=0\). Thus each set \(\{D>1/n\}\) is null, and \(\{D>0\}=\bigcup_{n\ge1}\{D>1/n\}\) is null. The converse follows because a function that vanishes a.e. has integral zero. By the equality condition in Young’s inequality, \(D=0\) a.e. exactly when \(A^p=B^q\) a.e. Substituting the definitions of \(A\) and \(B\) gives the claimed criterion in both directions.
If \(\|f\|_p=0\), then \(\int|f|^p\,d\mu=0\), so the zero-integral fact just proved gives \(f=0\) a.e. Consequently \(|fg|=0\) a.e., and both sides of Hölder’s inequality are zero. The same argument applies if \(\|g\|_q=0\). This also explains why the normalized criterion is reserved for the nonzero-norm case. \(\square\)
Worked Examples
Worked Example: Equality for Different Conjugate Exponents
On \([0,1]\) with Lebesgue measure \(m\), take \(p=3\), \(q=3/2\), \(f(x)=x\), and \(g(x)=x^2\). Check the equality criterion and compute both sides.
Both functions are measurable and belong to the required spaces, since $$ \|f\|_3^3=\int_0^1x^3\,dx=\frac14,\qquad \|g\|_{3/2}^{3/2}=\int_0^1(x^2)^{3/2}\,dx=\int_0^1x^3\,dx=\frac14. $$ Thus \(\|f\|_3=(1/4)^{1/3}\) and \(\|g\|_{3/2}=(1/4)^{2/3}\). Their normalized powers agree: for every \(x\in[0,1]\), $$ \frac{|f(x)|^3}{\|f\|_3^3}=\frac{x^3}{1/4}=4x^3 =\frac{x^3}{1/4} =\frac{|g(x)|^{3/2}}{\|g\|_{3/2}^{3/2}}. $$ The criterion therefore predicts equality. Direct computation checks it: $$ \int_0^1|f(x)g(x)|\,dx=\int_0^1x^3\,dx=\frac14, \qquad \|f\|_3\|g\|_{3/2}=(1/4)^{1/3}(1/4)^{2/3}=\frac14. $$ The two sides agree exactly. In particular, the power relation is \(|f|^3=|g|^{3/2}\), not \(|f|=|g|\).
Worked Example: A Strict Hölder Bound on a Two-Point Space
Let \(X=\{1,2\}\), let \(\mathcal{M}\) be all subsets of \(X\), and use counting measure \(\mu\). Set \(p=q=2\), \(f(1)=f(2)=1\), \(g(1)=1\), and \(g(2)=0\). Check that the bound is strict.
For counting measure, integrating a function means summing its values at the two points. Hence $$ \|f\|_2^2=1^2+1^2=2,\qquad \|f\|_2=\sqrt2, \qquad \|g\|_2^2=1^2+0^2=1,\qquad \|g\|_2=1. $$ The absolute product integral is $$ \int_X|fg|\,d\mu=|1\cdot1|+|1\cdot0|=1, $$ whereas the product of the norms is \(\sqrt2\cdot1=\sqrt2\). Since \(1<\sqrt2\), Hölder’s inequality is strict.
The criterion confirms the strictness. At the first point the normalized squares are \(1/2\) and \(1\), respectively; at the second they are \(1/2\) and \(0\). They are not equal at either point, and both points have positive counting measure. Therefore the required equality fails on a set of positive measure.
Check: the computed values satisfy \(1<\sqrt2\), and the normalized powers disagree point by point, as the equality criterion predicts.
Worked Example: Magnitude Equality but Cancellation in the Pairing
Again take \(X=\{1,2\}\) with counting measure and \(p=q=2\). Let \(f(1)=f(2)=1\) and \(g(1)=1,\ g(2)=-1\). Compare the absolute product integral with the modulus of the pairing \(\int f\overline g\,d\mu\).
At both points \(|f|=|g|=1\), so the normalized squares agree: \(|f|^2/\|f\|_2^2=|g|^2/\|g\|_2^2=1/2\). The norms are \(\|f\|_2=\sqrt{1^2+1^2}=\sqrt2\) and \(\|g\|_2=\sqrt{1^2+(-1)^2}=\sqrt2\). The absolute product integral is \(|1\cdot1|+|1\cdot(-1)|=2\), equal to the norm product \(\sqrt2\cdot\sqrt2=2\).
But the pairing itself is \(\int f\overline g\,d\mu=1\cdot1+1\cdot(-1)=0\), so its modulus is \(0\), not \(2\). The magnitudes attain Hölder’s bound, while the opposite signs cancel in the integral. These explicit sums verify both claims.
Equality for the Integral Pairing
For real- or complex-valued functions, Hölder also gives \(|\int f\overline g\,d\mu|\le\int|fg|\,d\mu\le\|f\|_p\|g\|_q\). Thus the criterion above fully describes equality in the second inequality. Equality in the combined bound \(|\int f\overline g\,d\mu|\le\|f\|_p\|g\|_q\) requires the magnitude condition and, when the product is nonzero, alignment of the complex phases.
To justify the phase condition, let \(h=f\overline g\), which is integrable. If \(|\int h\,d\mu|=\int|h|\,d\mu>0\), set \(\omega=(\int h\,d\mu)/|\int h\,d\mu|\). Then $$ \int\bigl(|h|-\operatorname{Re}(\overline{\omega}h)\bigr)\,d\mu =\int|h|\,d\mu-\left|\int h\,d\mu\right|=0. $$ The integrand is nonnegative, since \(\operatorname{Re}(\overline{\omega}h)\le|h|\). The zero-integral fact proved above implies \(\operatorname{Re}(\overline{\omega}h)=|h|\) a.e.; for a complex number this means \(h=\omega|h|\). Conversely, if \(h=\omega|h|\) a.e., then \(|\int h\,d\mu|=|\omega\int|h|\,d\mu|=\int|h|\,d\mu\). Combining this with the Hölder equality criterion proves the stated equivalence. If \(\int|h|=0\), then \(h=0\) a.e., and both pairing and product integral are zero.
A Common Mistake and Why the Distinction Matters
A common error is to conclude that \(|\int f\overline g\,d\mu|=\|f\|_p\|g\|_q\) as soon as the normalized powers agree. The two-point example with \(f=(1,1)\), \(g=(1,-1)\) gives a direct countercalculation: the normalized squares agree and the norm product is \(2\), but \(|\int f\overline g\,d\mu|=|1-1|=0\). The correction is to separate the two inequalities: proportional powers give equality in \(\int|fg|\,d\mu\le\|f\|_p\|g\|_q\); phase alignment is additionally needed to prevent cancellation in \(\int f\overline g\,d\mu\).
In the real-valued case, phase alignment reduces to a sign condition: on the set where \(fg\ne0\), the product must have one fixed sign a.e. for equality in the modulus of the pairing. For complex functions, its argument must be constant a.e. on that set. These conditions do not replace proportionality of the magnitudes; equality in the full pairing bound requires both.
The criterion is useful when deciding whether a Hölder estimate is sharp. It also identifies the shape that an equality case must have, rather than merely confirming a numerical bound. Always first check whether either norm is zero; if neither is zero, compare the normalized powers. For an integral without an absolute value around the product, then check whether signs or phases can cause cancellation.
Check Your Understanding
Use the normalized-power criterion for the absolute product integral, and check phase alignment separately when considering a pairing.
- For \(p=4\) and \(q=4/3\), write the normalized-power equality condition for nonzero \(f\in L^4(\mu)\) and \(g\in L^{4/3}(\mu)\).
- On a two-point space with counting measure, let \(f=(1,2)\), \(g=(1,2)\), and \(p=q=2\). Compute both norms and \(\int|fg|\,d\mu\), then check the equality criterion.
- If \(\|f\|_p=0\), explain why the normalized-power condition is not needed and why Hölder’s inequality is an equality.
- Give a phase condition that, together with proportional normalized powers, yields equality in \(|\int f\overline g\,d\mu|\le\|f\|_p\|g\|_q\).
- For \(f=(1,1)\), \(g=(1,-1)\) on a two-point counting-measure space with \(p=q=2\), compute \(\int|fg|\,d\mu\) and \(|\int f\overline g\,d\mu|\), and explain their difference.