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Equality and strictness in the uniform second-order remainder bound

Example

Two explicit one-variable cases of A uniform Hessian bound controls the second-order Taylor remainder.

A quadratic on an interval. Let $h (y) : = y^{2}$ on the open convex set $U = (- 1, 3)$ and fix $μ = 1$ . Then $h$ is $C^{2}$ with constant Hessian $(2)$ , so $∥ H_{h} (y) ∥ = 2$ for every $y \in U$ and $B = 2$ is the sharp constant. For every $y \in U$ , $h (y) - h (1) - h^{'} (1) (y - 1) = y^{2} - 1 - 2 (y - 1) = (y - 1)^{2} = \frac{B}{2} ∣ y - 1 ∣^{2},$ so the inequality of the lemma is an equality for every $y \in U$ .

A cubic on an interval. Let $g (y) : = y^{3}$ on the open convex set $V = (- 1, 2)$ and fix $μ = 1 / 2$ . Then $g$ is $C^{2}$ with Hessian $(6 y)$ , and $∥ H_{g} (y) ∥ = ∣ 6 y ∣ \leq 12$ on $V$ , so $B = 12$ is an admissible constant. At $y = 1$ , $g (1) - g (\frac{1}{2}) - g^{'} (\frac{1}{2}) (1 - \frac{1}{2}) = 1 - \frac{1}{8} - \frac{3}{4} \cdot \frac{1}{2} = \frac{1}{2} \leq \frac{12}{2} (\frac{1}{2})^{2} = \frac{3}{2},$ so the bound holds strictly at this point. Both computations are formal and assert nothing beyond the stated intervals.

Facts & Assumptions

Given: The functions $h (y) = y^{2}$ on $U = (- 1, 3)$ and $g (y) = y^{3}$ on $V = (- 1, 2)$ with the points indicated.

[L1]

If $U$ is open and convex, $h : U \to R$ is $C^{2}$ , and $∥ H_{h} (x) ∥ \leq B$ on $U$ , then $∣ h (y) - h (μ) - h^{'} (μ) (y - μ) ∣ \leq \frac{B}{2} ∣ y - μ ∣^{2}$ for all $μ, y \in U$ . (A uniform Hessian bound controls the second-order Taylor remainder)

[L2]

The derivatives of $y^{2}$ and $y^{3}$ are $2 y$ and $3 y^{2}$ , and the Hessian of a one-variable $C^{2}$ function is the $1 \times 1$ matrix of its second derivative. (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, The Hessian matrix and critical points of a scalar field, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case)

[L3]

For a $1 \times 1$ matrix $(a)$ the operator norm is $∣ a ∣$ . (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||)

[L4]

The intervals $(- 1, 3)$ and $(- 1, 2)$ are open and convex subsets of $R$ . (A convex subset of $R^{m}$ contains every line segment between two of its points)

Verification

technique · direct computation

1.1givenL2L3L4

On $U = (- 1, 3)$ the second derivative of $h$ is the constant $2$ , so $H_{h} = (2)$ and $∥ H_{h} ∥ = 2$ by [L2] and [L3]; $U$ is open and convex by [L4].

1.2givenL2L3L4algebra

On $V = (- 1, 2)$ the second derivative of $g$ is $6 y$ , so $H_{g} (y) = (6 y)$ and $∥ H_{g} (y) ∥ = ∣ 6 y ∣ \leq 12$ for $∣ y ∣ < 2$ ; $V$ is open and convex by [L4].

2.1step 1.1givenL1algebra

For $h$ with $μ = 1$ and $B = 2$ , the difference in the Example equals $(y - 1)^{2}$ by expansion, and the bound of [L1] is $\frac{2}{2} (y - 1)^{2}$ , so equality holds for every $y \in U$ .

2.2step 1.2givenL1algebra

For $g$ with $μ = 1 / 2$ and $B = 12$ , the difference at $y = 1$ is $1 - \frac{1}{8} - \frac{3}{8} = \frac{1}{2}$ , while the bound of [L1] is $\frac{12}{2} \cdot \frac{1}{4} = \frac{3}{2}$ , and $\frac{1}{2} < \frac{3}{2}$ .

3.1step 2.1step 2.2∎

Step 2.1 exhibits equality in the bound, and step 2.2 exhibits strict inequality, so both alternatives occur under the lemma's hypotheses.

Depends on

Dependency tree · 2 levels

Sources

K. Bryan, Taylor's Theorem in One and Several Variables (Rose-Hulman MA 433 lecture notes), the second-order case in n dimensions (equation (8)) and the Hessian matrix form 1/2 (x-a)^T H(b) (x-a)