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For bounded matrices $A_j$, the Lie product formula gives operator norm convergenceFor two summands, expanding their matrix exponentials gives the one-step difference below. For Hermitian matrices multiplied by $-i$, the factors are unitary operators and the first-order unitary product-formula error bound gives an explicit error estimate.

The matrix 2-norm is the operator norm induced by the Euclidean norm:It equals the largest singular value of $A$ and is invariant under multiplication by orthogonal or unitary matrices.

For $f:G\to M_n(\mathbb{C})$ and a degree-$d_{\rho }$ unitary irreducible representation, the matrix Fourier block is the operator $T_{\rho }$ on $M_{n\times d_{\rho }}(\mathbb{C})$ defined in the title. Its singular values ${\lambda }_{\rho ,j}$ satisfyPointwise operator norm at most one implies $\Vert T_{\rho }{\Vert }_{\mathrm{op}}\le 1$. These blocks are operators on a matrix space, rather than simply the entrywise scalar Fourier transform on a finite group without any reshaping.

An operator norm limit of compact operators between Banach spaces is compact. Approximate the image of the unit ball by the image under one compact operator and transfer a finite epsilon-net across the small operator-norm error.

For a real positive linear functional on the space of continuous functions on a compact space, $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ gives $|\varphi (f)|\le \varphi (1)\Vert f{\Vert }_{\infty }$. Evaluation at $1$ gives equality of the operator norm and $\varphi (1)$. Thus positivity alone already implies continuity in the supremum norm.

For a bounded linear operator on a complex Hilbert space, the numerical radius is the supremum of the absolute values in its numerical range of an operator. It is a norm equivalent to the operator norm, with $w(T)\le \Vert T\Vert \le 2w(T)$. For a self-adjoint operator the two norms agree.

If two closed subspaces of a Hilbert space give $H=V\oplus W$, the projection onto $V$ along $W$ fixes $V$ and vanishes on $W$. It need not be an orthogonal projection. The bounded inverse theorem applied to the addition map $(v,w)\mapsto v+w$ on the complete component spaces proves boundedness of the component maps. With both complementary subspaces nonzero, its operator norm and that of $I-Q$ equal $\sec {\theta }_{V,W^{\perp }}$, with the directed subspace angle convention. Indeed, for fixed $v\in V$, the smallest possible norm of $v+w$ over $w\in W$ is $\Vert P_{W^{\perp }}v\Vert$, which proves the formula for $\Vert Q\Vert$. Equality with $\Vert I-Q\Vert$ follows from the block representation $Q=\left(\begin{array}{cc}I& B\\ 0& 0\end{array}\right)$ on $V\oplus V^{\perp }$: both squared operator norms are $1+\Vert B{\Vert }^2$. If a summand is zero, $Q$ is zero or the identity operator, and one complementary operator norm is zero; those cases must be treated directly.

An orthogonal direct sum is a direct sum whose distinct summands are mutually orthogonal. For $v\in V$ and $w\in W$, the Pythagorean identity gives $\Vert v+w{\Vert }^2=\Vert v{\Vert }^2+\Vert w{\Vert }^2$. In a Hilbert space, a closed subspace of a Hilbert space and its orthogonal complement give an orthogonal direct sum of the entire space. The component maps are orthogonal projections and have operator norm one when their ranges are nonzero.

Suppose a piecewise finite difference uses two translated evaluations divided by $\alpha$. If the pulled-back evaluation intervals cover almost every input point at most $m$ times, then $|a-b{|}^2\le 2(|a{|}^2+|b{|}^2)$ gives $\Vert R_{\alpha }f{\Vert }_2^2\le 2m\Vert f{\Vert }_2^2/{\alpha }^2$. This converts an interval-overlap count into an operator norm estimate without requiring pointwise values of an L2 space equivalence class outside the almost-everywhere evaluation rule.