Saddle-Point (Laplace) Method in One Dimension

This lemma refines the Laplace principle by identifying the leading prefactor when the maximum is attained at a nondegenerate interior point.

Statement (one-dimensional nondegenerate maximum)

Let $f:[a,b]\to\mathbb{R}$ be twice continuously differentiable and suppose:

• $f$ has a unique global maximum at an interior point $x_0\in(a,b)$,
• $f''(x_0)<0$ (nondegenerate maximum),
• $g:[a,b]\to\mathbb{R}$ is continuous with $g(x_0)\neq 0$.

Define

Then, as $n\to\infty$,

Under higher smoothness assumptions on $f$ and $g$, one can obtain a full asymptotic expansion in powers of $1/n$.

Key hypotheses and conclusions

Hypotheses

• A unique interior maximizer $x_0$ for $f$.
• Nondegeneracy: $f''(x_0)<0$.
• Mild regularity of $g$ and $f$ near $x_0$.

Conclusions

• The integral is asymptotically Gaussian around $x_0$ after Taylor expansion of $f$: the leading exponential growth is $e^{n f(x_0)}$, and the subexponential prefactor is of order $n^{-1/2}$.
• This yields a quantitative refinement of the Laplace principle: $\frac1n\log I_n \to f(x_0)$.

Proof idea / significance

Expand $f$ near $x_0$: $f(x)=f(x_0)+\tfrac12 f''(x_0)(x-x_0)^2 + r(x)$, where $r(x)=o((x-x_0)^2)$. The main contribution to $I_n$ comes from a shrinking neighborhood of size $O(n^{-1/2})$ around $x_0$, on which $g(x)\approx g(x_0)$ and $r(x)$ is negligible compared to the quadratic term. After rescaling $y=\sqrt{n}(x-x_0)$, the integral reduces to a Gaussian integral, giving the stated prefactor. Contributions away from $x_0$ are exponentially smaller because $f(x)<f(x_0)$ there.

In statistical mechanics, saddle-point estimates justify mean-field and variational approximations of partition functions (for example, after introducing an order parameter or via Hubbard–Stratonovich transforms) and they often explain the appearance of Gaussian fluctuations around equilibrium points.