Log-Sobolev inequalities and Herbst's argument
Derive a centered exponential-moment bound from a precisely normalized entropy–energy inequality and justify the regularity steps.
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The question for today
What you’ll understand
- State a log-Sobolev inequality with an explicit constant convention.
- Derive the Herbst differential inequality and its integrated mgf bound.
- Track gradient and scaling constants while distinguishing an argument from an unproved regularity assumption.
The Gaussian theorem fits a broader mechanism: an entropy–energy inequality can force every sufficiently regular Lipschitz observable to have a quadratic log mgf. Herbst's argument reveals this connection by applying the inequality to an exponential test function. Factors of two move between the squared test function, its gradient, and the centered mgf, so declaring the normalization is essential. The differentiations are straightforward for bounded smooth functions; extending beyond them is a separate analytic step.
Declare the entropy and constant
For a nonnegative integrable Y with the required logarithmic integrability, define Ent_μ(Y)=E_μ[Y log Y]−E_μY log(E_μY), with zero log zero interpreted as zero. Say a probability measure μ on Euclidean space satisfies a log-Sobolev inequality with constant C>0 when Ent_μ(g²)≤2C E_μ||∇g||₂² for admissible real test functions. We initially assume the inequality for bounded continuously differentiable g with the finite energy needed below, and its usual extension by approximation when invoked.
Under this convention the standard Gaussian has C=1, while covariance σ²I has C=σ². Sources that place C rather than 2C before the energy use a different numerical constant. The gradient is Euclidean; changing the metric also changes the energy and its constant. The inequality compares a nonlinear entropy with a quadratic local energy, and it is stronger than merely controlling variance. Expanding near a constant recovers a Poincaré-type variance bound, but that weaker bound alone does not reproduce the full exponential-moment conclusion.
The exponential test creates a differential inequality
First let f be bounded and continuously differentiable with gradient norm at most L, and center it by subtracting its expectation. Define H(λ)=log E_μ exp(λ(f−E_μf)). Boundedness justifies differentiating under the expectation for every finite real λ. Insert g=exp(λ(f−E_μf)/2) into the log-Sobolev inequality. The squared gradient supplies λ²/4, while the left entropy, divided by the exponential expectation, becomes λH'(λ)−H(λ).
The result bounds λH'−H by Cλ²L²/2. For positive λ, divide by λ² to control the derivative of H(λ)/λ. Its limit at zero is zero because f was centered, so integration gives H(λ)≤CL²λ²/2. Apply the same calculation to −f for negative λ. Chernoff then yields a one-sided tail exp(−t²/(2CL²)) and its two-sided factor-two version. The argument is an explicit derivation for the stated test-function class, not a general license to differentiate any infinite exponential moment.
Extend the function class carefully
For an integrable globally Lipschitz f, truncate its values and smooth by a Euclidean mollifier. The approximants can retain Lipschitz constant at most L, converge pointwise, and have expectations converging to Ef; the latter follows from integrability and controlled approximation. Assuming the log-Sobolev inequality extends to these bounded smooth tests, the derived mgf bounds are uniform. Fatou's lemma then passes the bound to exp(λ(f−Ef)), establishing finiteness rather than assuming it. This explains how an unbounded linear Gaussian observable can be included rigorously.
Entropy dissipation under a suitable Markov semigroup is another route to proving a log-Sobolev inequality, but a decaying mode by itself does not establish that inequality for every test function. A periodic heat equation and a Gaussian Ornstein–Uhlenbeck semigroup use different invariant measures and generators. They share an energy-dissipation idea without being the same model. The lab therefore illustrates mode decay only; the concentration constants here come from the stated entropy inequality and its verified Gaussian scaling, not from fitting a decay curve.
Work it through
Check the factors on a scaled Gaussian
Let μ=N(0,4I_d), with log-Sobolev constant C=4 under the stated convention, and take f(x)=x₁. Verify the Herbst proxy directly and bound P(f−Ef≥4).
The Euclidean gradient of f is the first unit vector, so L=1 and Ef=0. Its centered log mgf is H(λ)=2λ² because the coordinate variance is four. The unbounded test is covered by the approximation extension, or checked directly by this finite Gaussian integral.
Differentiate: λH'(λ)−H(λ)=λ·4λ−2λ²=2λ². The log-Sobolev right-hand side after dividing by the exponential expectation is CL²λ²/2=2λ², so this example attains equality in the differential bound.
The proxy is CL²=4. At threshold four, the directed tail bound is exp(−16/(2·4))=exp(−2)≈0.135335. A two-sided event would have factor two; the exact one-sided Gaussian probability is smaller.
The declared normalization gives proxy four and the one-sided bound exp(−2). The direct mgf verifies every factor of two.
The living notebook
Explore the model
Interactive lab · heat
Mode decay as an energy-dissipation analogy
Pause and Reset, then set mode n = 1 and diffusivity κ = 0.5. After two Steps, time is one and amplitude is exp(−0.5)≈0.606531. Reset and set both n = 2 and κ = 0.5, then take two Steps: amplitude is exp(−2)≈0.135335. The displayed periodic heat model has decay κn², not a Gaussian Ornstein–Uhlenbeck generator, and computes no entropy or log-Sobolev constant. Use it only to compare the rates at which these two modes dissipate.
- Current amplitude
- 1
- Spatial L² norm
- 1.772
- Decay rate · κn²
- 3.6 /s
The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forObserved decay motivates an entropy–energy mechanism, but does not prove the full inequality or identify its measure-dependent constant.
Pause & reason
Check your understanding
Your turn
Try it yourself
A measure satisfies the stated log-Sobolev inequality with C=3, and an admissible integrable function is globally 2-Lipschitz. Give the upper-tail and two-sided bounds for deviation six. State what must justify an unbounded function's inclusion.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
First compute CL². For an unbounded function, describe the bounded smooth approximation and the passage to its exponential moments.
Compare with the solution
The proxy is CL²=3·4=12. The upper-tail bound is and the two-sided bound is . For unbounded f, one needs integrability, bounded smooth approximants with the same Lipschitz bound, validity of the entropy inequality on those tests, and convergence of their centers. Fatou's lemma then transfers the uniform exponential bound. Merely writing down H and differentiating it before knowing its expectation is finite would leave a gap that this approximation argument resolves.
Keep the idea
Take it with you
- The placement of 2C in the entropy–energy inequality fixes the concentration constant.
- Herbst turns an exponential test into a differential bound for the centered log mgf.
- Bounded smooth derivations extend to integrable Lipschitz functions through justified approximation and Fatou's lemma.
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