Young's Inequality

The statement and the case p = q = 2

For a, b ≥ 0 and conjugate exponents p, q > 1 with 1/p + 1/q = 1:

ab ≤ a^p / p + b^q / q

Equality holds iff a^p = b^q. The case p = q = 2 (1/2 + 1/2 = 1) gives the cleanest specialisation:

ab ≤ a²/2 + b²/2          (rearranges to (a − b)² ≥ 0)

And the parameterized version with ε > 0 (also commonly called "Young's inequality with ε"):

ab ≤ ε a^p / p + ε^(−q/p) b^q / q       (rescale a → ε^(1/p) a, b → ε^(−1/p) b)

The ε version is the everyday workhorse in PDE estimates: it lets you absorb a small piece of a cross-term ab into εa^p/p (a small fraction of the dominant energy term) while paying with the rest as ε^(−q/p) b^q/q.

Proof via concavity of log

For a, b > 0 (the case a = 0 or b = 0 is trivial), use the weighted average:

log(ab) = log a + log b = (1/p) · p log a + (1/q) · q log b
        = (1/p) log(a^p) + (1/q) log(b^q)

Since log : (0, ∞) → ℝ is strictly concave, Jensen gives

(1/p) log(a^p) + (1/q) log(b^q) ≤ log( (1/p) a^p + (1/q) b^q )

Combine and exponentiate: ab ≤ (1/p) a^p + (1/q) b^q. Equality holds iff a^p = b^q (the two arguments of log agree). The proof is exactly weighted AM-GM with weights 1/p, 1/q applied to x₁ = a^p, x₂ = b^q. Young is weighted AM-GM in disguise.

Geometric proof — the rectangle and the curve

Consider the curve y = x^(p−1) in the first quadrant. Inverting: x = y^(1/(p−1)) = y^(q−1), where q = p/(p−1). The curve passes through the origin, is increasing, and divides the first quadrant into two regions.

Pick any point (a, b) in the first quadrant. The rectangle [0, a] × [0, b] has area ab. The area under the curve from 0 to a is:

∫₀ᵃ x^(p−1) dx = a^p / p

The area to the left of the curve from 0 to b (integrating the inverse) is:

∫₀ᵇ y^(q−1) dy = b^q / q

If (a, b) lies on the curve, i.e., b = a^(p−1) ⟺ a^p = b^q, the rectangle is tiled exactly by these two regions; their areas sum to ab. If (a, b) lies off the curve (above or below), the two regions either overlap or leave a gap — either way their combined area is at least ab. So:

ab ≤ a^p / p + b^q / q

with equality exactly when (a, b) is on the curve. This is the picture every textbook draws.

Worked numerical examples

Example 1 (p = q = 2):
  a = 3, b = 4
  ab = 12
  a²/2 + b²/2 = 4.5 + 8 = 12.5
  12 ≤ 12.5            ✓  (slack = (3−4)²/2 = 0.5)

Example 2 (p = q = 2, equality):
  a = b = 5
  ab = 25
  a²/2 + b²/2 = 12.5 + 12.5 = 25
  25 = 25              EQUALITY (a² = b²)

Example 3 (p = 3, q = 3/2):
  a = 2, b = 4
  ab = 8
  a^p/p = 8/3 ≈ 2.667
  b^q/q = 4^(3/2) / (3/2) = 8 / 1.5 ≈ 5.333
  a^p/p + b^q/q ≈ 8.000
  8 ≤ 8.000            EQUALITY (since a^p = 2³ = 8 = 4^(3/2) = b^q)

Example 4 (p = 4, q = 4/3):
  a = 1, b = 2
  ab = 2
  a^p/p = 1/4 = 0.25
  b^q/q = 2^(4/3) / (4/3) ≈ 1.890
  a^p/p + b^q/q ≈ 2.140
  2 ≤ 2.140            ✓  (slack ≈ 0.140)

Example 5 (parameterized Young, ε = 0.1, p = q = 2):
  a = 5, b = 5
  ab = 25
  ε a²/2 + ε^(−1) b²/2 = 0.1·12.5 + 10·12.5 = 1.25 + 125 = 126.25
  25 ≤ 126.25          ✓  (the ε trick gives a very loose bound when ε is small —
                             but it lets you put almost all the cost in one term)

Variants and generalizations

• Scalar Young. ab ≤ a^p/p + b^q/q for 1/p + 1/q = 1.
• Parameterized Young with ε. ab ≤ ε a^p/p + ε^(−q/p) b^q/q. Standard in PDE estimates — tune ε to absorb cross-terms into the principal energy.
• Fenchel-Young (general convex). For a convex function φ with Legendre dual φ*, ab ≤ φ(a) + φ*(b). Setting φ(x) = x^p/p recovers scalar Young. The general statement is the duality inequality of convex analysis.
• Young's convolution inequality. ‖f * g‖_r ≤ ‖f‖_p · ‖g‖_q for 1 ≤ p, q, r ≤ ∞ with 1/p + 1/q = 1 + 1/r. Sharp constant (Beckner-Brascamp-Lieb 1975) attained by Gaussians.
• Matrix Young. For positive semi-definite matrices A, B: ‖AB‖ ≤ (‖A‖^p)/p + (‖B‖^q)/q in operator norm, with appropriate non-commutative subtleties.
• Reverse Young. For 0 < p < 1 (so q < 0) and a, b > 0, the inequality reverses: ab ≥ a^p/p + b^q/q. Used in reverse Hölder / Gehring's lemma machinery.
• Generalized n-term Young. For weights wᵢ summing to 1 and aᵢ ≥ 0: Πaᵢ^{wᵢ} ≤ Σwᵢ aᵢ (this is exactly weighted AM-GM).

Young proves Hölder (one application)

Apply Young pointwise to a = |f(x)| / ‖f‖_p, b = |g(x)| / ‖g‖_q. Then

|f(x) g(x)| / (‖f‖_p · ‖g‖_q) ≤ (1/p) |f(x)|^p / ‖f‖_p^p + (1/q) |g(x)|^q / ‖g‖_q^q

Integrate both sides over the measure space. The right side becomes 1/p + 1/q = 1. Multiplying both sides by ‖f‖_p · ‖g‖_q:

∫ |f g| dμ ≤ ‖f‖_p · ‖g‖_q          (Hölder's inequality)

One pointwise application of Young, one integration. The hierarchy convexity ⇒ Young ⇒ Hölder ⇒ Minkowski is tight: each step is short, and the cumulative power produces all of L^p theory.

The Legendre/Fenchel duality view

For a convex function φ : ℝ → ℝ ∪ {+∞}, its Legendre conjugate is φ*(y) = sup_x (xy − φ(x)). Equivalently, φ* is the convex function such that y ↦ x is the inverse derivative relationship. The Fenchel-Young inequality says, for all a, b:

ab ≤ φ(a) + φ*(b)

with equality iff b ∈ ∂φ(a) (the subgradient of φ at a). For φ(x) = x^p/p on [0, ∞), the Legendre dual is φ*(y) = y^q/q on [0, ∞), with q = p/(p−1) and the derivative relation y = x^(p−1) ⟺ x = y^(q−1). Substituting recovers scalar Young.

This view explains why Young's inequality appears across thermodynamics (energy/entropy duality), large-deviations theory (Cramér's theorem), and convex optimization (Lagrangian duality, primal/dual gap). Young is the prototypical Fenchel-Young inequality.

Common pitfalls

• Forgetting non-negativity. Young requires a, b ≥ 0 (or absolute values for general reals). With negative values, the powers may not be defined or the inequality may flip.
• Forgetting the conjugate condition. 1/p + 1/q = 1 is mandatory. With arbitrary p, q the constants on the right side don't add to 1 and the bound is wrong.
• Confusing scalar and convolutional Young. The scalar form bounds ab pointwise; the convolutional form bounds the L^r norm of f * g. Both are called "Young", different statements.
• Using a = b case as a general pattern. The case p = q = 2 is the simplest but not the most useful. The non-Cauchy-Schwarz cases (p = 3, q = 3/2; p = 4, q = 4/3) are where Young's flexibility matters.
• Forgetting the ε version's purpose. Use it to absorb cross-terms when one factor is small. The cost is that the other factor blows up — there is no free lunch.
• Believing the Legendre proof is "harder". Once you know the Legendre dual of x^p/p, the inequality is the defining property of the conjugate. The unfamiliarity comes from the language, not the depth.

Where Young's inequality shows up

• L^p theory. Young → Hölder → Minkowski → triangle inequality for ‖·‖_p → L^p is a normed space. Half of functional analysis rests on this chain.
• PDE energy estimates. Bound a quadratic cross-term in an energy identity by Young's inequality with ε, absorb the εa^p/p part into the principal energy, control the rest. Routine in Navier-Stokes, NLS, KdV, and parabolic regularity theory.
• Sobolev embedding. Bounds like ‖f‖_q ≤ C(p, q) ‖∇f‖_p (Gagliardo-Nirenberg-Sobolev) use Hölder, which rests on Young.
• Optimal transport and Wasserstein. Kantorovich-Rubinstein duality, optimal cost-density tradeoffs, and the Brenier theorem all use Legendre/Fenchel-Young.
• Statistical mechanics / large deviations. Cramér's theorem expresses log-moment-generating function as Legendre dual of rate function — Young's inequality is the duality bound that proves the rate function is convex.
• Information theory. Donsker-Varadhan variational formula for KL divergence; chi-squared and KL bounds on hypothesis testing all use Young/Fenchel-Young.
• Optimization — duality. Lagrange duality: weak-duality bound and strong-duality conditions (Slater, KKT) rest on the Fenchel-Young inequality applied to the Lagrangian.
• Harmonic analysis. Young's convolution inequality is the standard tool for bounding convolutional smoothing operators. Sharp constants involve Gaussians (Beckner, Brascamp-Lieb).