A uniform Hessian bound controls the second-order Taylor remainder

Given: An open convex set $U\subseteq {\mathbb{R}}^m$, a $C^2$ function $h:U\to \mathbb{R}$, a constant $B\ge 0$ with $\parallel H_h(x)\parallel \le B$ on $U$, and points $\mu ,y\in U$.

For $f\in C^2(U)$ on open convex $U$ and $a,a+h\in U$, the Lagrange remainder formula gives some $\theta \in (0,1)$ with $f(a+h)=T_{1}f(a;h)+{\sum }_{\mid \alpha \mid =2}\frac{D^{\alpha }f(a+\theta h)}{\alpha !}{h}^{\alpha }\mathrm{.}$ (Multivariable Taylor formula with a Lagrange remainder along a line segment)

The degree-one Taylor polynomial is $T_{1}f(a;h)=f(a)+{\sum }_{j=0}^{m-1}{\mathrm{\partial }}_{j}f(a)h_j=f(a)+\mathrm{\nabla }f(a)\cdot h$, using the zero-based coordinates of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case and the multi-index convention of The multivariable Taylor polynomial in multi-index notation.

Since $f\in C^2(U)$, its mixed partial derivatives commute by Clairaut–Schwarz (Clairaut--Schwarz theorem for continuous second partial derivatives). Thus the second-order multi-index sum of [L1] equals $\frac{1}{2}{v}^{\mathsf{T}}{H}_f(a+\theta h)v$ for $v=h$: the diagonal terms are $\frac{1}{2}({\mathrm{\partial }}_i^{2}f)v_i^2$ and each off-diagonal pair contributes $({\mathrm{\partial }}_i{\mathrm{\partial }}_{j}f)v_i{v}_j$. The Hessian convention is $H_f=({\mathrm{\partial }}_i{\mathrm{\partial }}_{j}f{)}_{i,j<m}$ (The Hessian matrix and critical points of a scalar field).

For a real $m\times m$ matrix $A$ and $v\in {\mathbb{R}}^m$, $\mid v^{\mathsf{T}}Av\mid \le \parallel Av\parallel \parallel v\parallel \le \parallel A\parallel \parallel v{\parallel }^2$, since $\mid ⟨Av,v⟩\mid \le \parallel Av\parallel \parallel v\parallel$ and $\parallel Au\parallel \le \parallel A\parallel$ for unit $u$. (Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs, The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||)

For a linear map between finite-dimensional inner-product spaces with nonzero domain, the operator-norm maximum exists, is attained, and equals the largest singular value. For the zero domain the norm is defined as $0$, with no unit vector at which to attain a maximum. Here the domain is ${\mathbb{R}}^m$ with $m\ge 1$. (The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector, The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||)

Matrices in $M_{m\times m}(\mathbb{R})$ have zero-based entry indices $0\le i,j<m$ (The vector space $M_{m\times n}(F):=F^{m\times n}$ of $m$ by $n$ matrices over a field, with entrywise operations). A map $L:{\mathbb{R}}^m\to {\mathbb{R}}^m$ is linear when $L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$ for all $u,v\in {\mathbb{R}}^m$ and $\alpha ,\beta \in \mathbb{R}$; the coordinate formula for $\widehat{M}$ in the Statement satisfies this identity termwise. (A linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^n$ in Euclidean coordinates)