Young's Inequality For Products

The Scalar Inequality and Its Setting

Young’s inequality is a pointwise bound for a product. It is one of the basic scalar tools behind inequalities for integrals and $L^p$ norms, but those later applications are not needed to prove it. The statement here is purely scalar, so it does not require a measure space.

Throughout, $p$ is a real number with $1<p<\infty$, and $q$ is its conjugate exponent, defined by $1/p+1/q=1$. Equivalently, $q=p/(p-1)$, so $1<q<\infty$. The variables $a$ and $b$ are finite nonnegative real numbers. These restrictions matter: the formula uses both finite powers and the positive denominators $p$ and $q$.

Theorem (Young’s inequality for products): If $1<p<\infty$, $q=p/(p-1)$, and $a,b\ge0$, then $$ ab\le \frac{a^p}{p}+\frac{b^q}{q}. $$ Equality holds if and only if $a^p=b^q$. This includes $a=b=0$; if exactly one of $a,b$ is zero, the inequality is strict.

The conclusion gives a sum of two powers in place of a product. The conjugacy condition makes the coefficients add to one, which is what permits the product to be bounded with these particular constants. The equality condition is a relation between powers, not generally the simpler-looking condition $a=b$.

Proof by Finding the Minimum

Fix $b\ge0$ and consider the real-valued function on $[0,\infty)$

F(a)=\frac{a^p}{p}-ab+\frac{b^q}{q}.

Showing Young’s inequality is equivalent to showing $F(a)\ge0$. The cases $b=0$ and $b>0$ are handled separately so that no division or derivative argument hides a boundary case.

Proof. If $b=0$, then $F(a)=a^p/p\ge0$ for every $a\ge0$. It equals zero exactly when $a=0$. The inequality therefore holds in this case, with equality exactly when $a=b=0$.

Now suppose $b>0$. For $a>0$, differentiation gives $F'(a)=a^{p-1}-b$. Since $p-1>0$, the function $a\mapsto a^{p-1}$ is strictly increasing on $(0,\infty)$. Thus $F'(a)<0$ when $0<a<b^{1/(p-1)}$, $F'(a)=0$ when $a=b^{1/(p-1)}$, and $F'(a)>0$ when $a>b^{1/(p-1)}$. By the derivative sign, $F$ decreases up to that point and increases after it. Its minimum on $[0,\infty)$ is therefore attained at $a_0=b^{1/(p-1)}$, also accounting for the endpoint $a=0$.

Because $q=p/(p-1)$, we have $a_0^p=b^q$ and $a_0b=b^{1+1/(p-1)}=b^q$. Hence

F(a_0)=\frac{b^q}{p}-b^q+\frac{b^q}{q} =b^q\left(\frac1p+\frac1q-1\right)=0.

It follows that $F(a)\ge F(a_0)=0$ for every $a\ge0$, proving the inequality. For $b>0$, the decrease-then-increase argument shows the minimum is attained only at $a_0$, so equality holds exactly when $a=b^{1/(p-1)}$. Raising both sides to the positive power $p$ gives the equivalent condition $a^p=b^q$. Together with the $b=0$ case, this proves both directions of the stated equality criterion. $\square$

This proof also identifies the equality balance directly: for fixed $b>0$, the product term is balanced against the $a^p/p$ term at the unique minimum, and the conjugate exponent makes the minimum exactly zero. No assumption about integrability or a measure space is involved.

Worked Scalar Calculations

Worked Example: The Quadratic Case

Take $p=2$, so $q=2/(2-1)=2$. For $a=3$ and $b=1$, Young’s inequality gives

ab=3\cdot1=3,\qquad \frac{a^p}{p}+\frac{b^q}{q} =\frac{3^2}{2}+\frac{1^2}{2} =\frac92+\frac12=5.

Thus $3\le5$, and the inequality is strict. The equality test gives $a^p=3^2=9$, whereas $b^q=1^2=1$, so the powers are not equal. Check: subtracting the product from the right side gives $5-3=2>0$, consistent with strict inequality and with the failed equality condition.

Worked Example: Equality for Unequal Inputs

Take $p=3$. Its conjugate exponent is $q=3/(3-1)=3/2$. Let $a=2$ and $b=4$. The product is $ab=2\cdot4=8$, and the two terms on the right are

\frac{a^p}{p}=\frac{2^3}{3}=\frac83,\qquad \frac{b^q}{q}=\frac{4^{3/2}}{3/2}=\frac{8}{3/2}=\frac{16}{3}.

The right side is $8/3+16/3=24/3=8$, so equality holds. The criterion confirms this: $a^p=2^3=8$ and $b^q=4^{3/2}=(\sqrt4)^3=2^3=8$. Check: both the product and the sum of the two terms equal $8$. The inputs are unequal, showing why equality should be stated as $a^p=b^q$, not $a=b$.

Worked Example: A Strict Bound with Fractional Powers

Take $p=3/2$, so $q=(3/2)/(1/2)=3$, and set $a=2$, $b=1$. Then $ab=2$. The right side is

\frac{a^p}{p}+\frac{b^q}{q} =\frac{2^{3/2}}{3/2}+\frac{1^3}{3} =\frac{4\sqrt2}{3}+\frac13 =\frac{4\sqrt2+1}{3}.

To verify the comparison exactly, $ (4\sqrt2+1)/3>2$ is equivalent to $4\sqrt2>5$. Both sides of the latter inequality are positive, and their squares satisfy $32>25$. Hence the right side is strictly greater than $2=ab$. The equality test also predicts strictness: $a^p=2^{3/2}=2\sqrt2$, while $b^q=1^3=1$, so $a^p\ne b^q$. Check: the exact comparison and the equality criterion give the same conclusion.

Worked Example: A Zero Input

Take $p=q=2$, $a=0$, and $b=3$. The product is $ab=0$. The right side is

\frac{a^p}{p}+\frac{b^q}{q} =\frac{0^2}{2}+\frac{3^2}{2} =0+\frac92=\frac92.

Thus $0\le9/2$, strictly. The equality test gives $a^p=0$ and $b^q=9$, which are different. This computation illustrates why the zero case must not be omitted: when exactly one input vanishes, the product is zero but the right side is positive. Check: $9/2>0=ab$, as required.

Real and Complex Inputs

The scalar theorem can be stated for real or complex numbers by applying it to their magnitudes. This extension does not require new assumptions on a measure space; it is still a pointwise statement about two numbers.

Corollary (Magnitude form): For real or complex numbers $x,y$ and conjugate exponents $1<p<\infty$, $q=p/(p-1)$, $$ |xy|\le\frac{|x|^p}{p}+\frac{|y|^q}{q}. $$ Equality holds if and only if $|x|^p=|y|^q$.

Proof. Set $a=|x|$ and $b=|y|$. These are nonnegative real numbers, so Young’s inequality gives $ab\le a^p/p+b^q/q$. The multiplicative property of absolute value gives $|xy|=|x||y|=ab$, proving the displayed bound. Its equality condition is $a^p=b^q$, which is exactly $|x|^p=|y|^q$. $\square$

For real numbers, the non-absolute product $xy$ is also bounded above by the same right side, since $xy\le |xy|$. Equality in this signed bound requires both $|x|^p=|y|^q$ and $xy=|xy|$. In particular, if the magnitudes are nonzero, the signs must agree; if both numbers are zero, equality holds as well.

Why the Conjugate Exponent and Constants Matter

The coefficients $1/p$ and $1/q$ are matched to the exponents. The proof used $1/p+1/q=1$ at the minimum to make the expression vanish. If the exponents are not conjugate, this argument does not produce Young’s inequality as stated, and the formula can fail.

A common mistake is to use $p=2$ and $q=3$ as though they were conjugate. At $a=b=1$, that incorrect substitution would claim

1=ab\le\frac{1^2}{2}+\frac{1^3}{3} =\frac12+\frac13=\frac56,

which is false because $1>5/6$. The correction is to check conjugacy first: if $p=2$, then $q=p/(p-1)=2$, and the valid inequality at $a=b=1$ is $1\le1/2+1/2=1$, with equality. The coefficients are not decorations that can be paired arbitrarily with powers.

Another pitfall is to assume equality means the two inputs are equal. The equality condition is $a^p=b^q$. For $p=3$ and $q=3/2$, the values $a=2$ and $b=4$ satisfy $a^p=b^q=8$, although $a\ne b$. In applications where the inputs are functions, Young’s inequality can be applied at each point where their values are defined; any subsequent integration requires its own hypotheses and justification. The scalar inequality alone does not assert that products of measurable functions are integrable.

The usefulness of Young’s inequality is that it converts a product into separate powers with explicit coefficients. Its proof is short, but the exact conjugacy requirement and equality condition are essential when the inequality is later used to control products.

Check Your Understanding

For each question, first check the range of the exponents and then use the equality criterion or the one-variable function from the proof.

For $p=4$, compute its conjugate exponent $q$, and determine whether equality holds at $a=2$, $b=8$.
For conjugate $p,q$, show directly from the equality criterion that $a=0$, $b>0$ gives strict inequality.
Use the magnitude form to write Young’s inequality for complex numbers $z,w$.
Explain why the derivative $F'(a)=a^{p-1}-b$ has exactly one zero when $b>0$.
At $a=b=1$, explain why choosing exponents that are not conjugate need not give a valid bound.