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Let $A_h:X\to Z$, $0<h<1$, be bounded linear operators from a normed space into a Banach space, with $M={\sup }_h\Vert A_h\Vert <\infty$. If $A_{h}d$ converges as $h\to 0^{+}$ for each element of a dense subspace $D$, then $A_{h}x$ converges for every $x\in X$. The limit defines a bounded linear operator $A$ with $\Vert A\Vert \le M$.

For $d\in D$, the triangle inequality givesApproximate $x$ by $d$ and use the Cauchy criterion for a function limit on $D$. The completeness of $Z$ supplies the limit; passing to the limit preserves linearity and the bound. When every limit on $D$ is zero, the inequality $\Vert A_{h}x\Vert \le M\Vert x-d\Vert +\Vert A_{h}d\Vert$ shows $A=0$.

A dense subspace of a topological vector space $X$ is a linear subspace $D$ with $\overline{D}=X$. For a normed vector space, every vector of $X$ is a limit of a sequence from $D$. A continuous linear operator into a Hausdorff target is uniquely determined by its values on $D$.

Existence of an extension needs additional hypotheses. A bounded linear operator from $D$ into a Banach space extends uniquely to $X$ by taking limits, as in extension of a bounded linear operator from a dense subspace. An incomplete target can obstruct this construction, as the failure of dense-subspace extension into an incomplete codomain shows.

For a finite interval $(a,b)$ with Lebesgue measure, restrictions of continuous functions on $[a,b]$ form a dense subspace of the Lp space $L^p((a,b))$ for $1\le p<\infty$. One proof first approximates by simple functions. The regularity of Lebesgue measure approximates each of their measurable sets in measure by a finite union of intervals. An interval's indicator function is approximated by continuous functions taking values in $[0,1]$ and differing from that indicator function only in endpoint strips of total length tending to zero. Their Lp norm error is at most the $1/p$ power of that length. The triangle inequality completes the approximation of each simple function.

If a normed subspace $Z$ is dense in a normed vector space $X$ and carries the inherited norm, restriction gives an onto linear isometry from the dual space of $X$ to that of $Z$. A bounded functional on $Z$ extends uniquely by limits along sequences from $Z$; its norm is unchanged because its inequality passes to the limit. Thus completion does not change the continuous dual. For example, the subspace of sequences with $\sum j|a_j|<\infty$, equipped with the unweighted ${\ell }^1$ norm, has dual ${\ell }^{\infty }$, since it contains the dense subspace of finitely supported sequences. The inherited norm matters; the weighted norm would give a different dual norm.

Suppose linear operators $T_n$ on a normed vector space satisfy $\Vert T_n\Vert \le C$ uniformly and $T_{n}p\to p$ on a dense subspace. The displayed triangle inequality for any element $p$ of that subspace first permits $p$ to approximate $f$, then lets $n$ increase, proving $T_{n}f\to f$ for every $f$. Applied to the Bernstein polynomial, the Weierstrass approximation theorem supplies the dense polynomial subspace and $C=1$.

Retain the first $N\ge 2$ number states of a quantum harmonic oscillator. Let ${\Pi }_N$ be the orthogonal projection onto their span and compress $X_N={\Pi }_{N}X{\Pi }_N$, $P_N={\Pi }_{N}P{\Pi }_N$, restricted to this span. The compressed annihilation operator $a_N$ satisfiesFor $n<N-1$, the two products give $n+1$ and $n$; on the top state they give $0$ and $N-1$. Thus, with $\ell =\sqrt{\hslash /(m\omega )}$,On every quantum superposition of $|0⟩,\dots ,|N-2⟩$, this commutator acts exactly as $i\hslash I_N$. For a normalized retained state with coefficients $c_n$,Small probability at the cutoff must therefore be judged relative to $1/N$ for this expectation. For $N=8$, the diagonal of $[X_N,P_N]/(i\hslash )$ is $(1,1,1,1,1,1,1,-7)$, with zero trace.

This is unavoidable for finite matrices: trace of a matrix commutator gives $\mathrm{tr}(XP-PX)=0$, whereas $\mathrm{tr}(i\hslash I_N)=i\hslash N\ne 0$. Compression does not preserve products: ${\Pi }_{N}XP{\Pi }_N$ can use intermediate states outside the retained span, while $X_N{P}_N$ cannot. Accordingly, compressing the full oscillator Hamiltonian operator gives ${\Pi }_{N}H{\Pi }_N=\hslash \omega (a_N^{†}{a}_N+I_N/2)$, but building it from the compressed squares givesThe two prescriptions agree on lower number states and disagree at the boundary. In the infinite oscillator, the canonical commutation relation holds on the common invariant dense subspace of finite number state combinations. Its unbounded position and momentum operators do not admit the finite trace argument; writing the formal identity does not assert that both products act on every vector of the Hilbert space.

The orthogonal complement is $W^{\perp }=\{v\in V:⟨v,w⟩=0\text{ for every }w\in W\}$. Put $Pv={\sum }_i⟨v,e_i⟩e_i$, so $T=I-P$. Orthonormality gives $P^2=P$, and therefore $T^2=T$. Also $⟨Tv,e_j⟩=0$ for every $j$, so $Tv\in W^{\perp }$. Every $v=Pv+Tv$ is thus a sum from $W$ and $W^{\perp }$, and their intersection is zero, since a vector orthogonal to itself has squared norm zero. This proves $V=W\oplus W^{\perp }$. The decomposition is unique, so its two projection maps depend only on $W$, not on the chosen orthonormal basis.

Finite dimensionality of $W$ cannot simply be dropped. Take the real Hilbert space $V={\ell }^2(\mathbb{N};\mathbb{R})$ and $W=c_{00}$, the finitely supported sequences. This is an infinite-dimensional proper subspace. A vector orthogonal to every coordinate unit vector is zero, so $W^{\perp }=\{0\}$. Yet, for example, $(2^{-j}{)}_{j\ge 1}$ belongs to $V\setminus W$. Hence $W\oplus W^{\perp }=W\ne V$. The failure occurs even though $V$ itself is complete; the proper dense subspace $W$ is not closed.

Use $D=C([-1,1])$, restricted to $(-1,1)$ and regarded as a linear subspace of $L^p((-1,1))$. The density of continuous functions in Lp on an interval says that $D$ is a dense subspace whenever $1\le p<\infty$.

For $g\in D$, the supremum norm givesIn particular $D\subseteq Y$, with zero limit. Part (c), together with $D$ being a dense subspace, implies $Y=X$. To identify the limit throughout $X$, for $f\in X$ and $g\in D$ useFirst choose $g$ in the dense subspace to make the first term arbitrarily small, then let $h\to 0^{+}$. This is convergence on a dense subspace under a uniform operator bound, and provesThe zero linear functional belongs to the continuous dual space $X^{\ast }$.

The Weierstrass approximation theorem states that the polynomials are a dense subspace of $C[0,1]$ in the supremum norm. By linearity and (c), $B_{n}p\to p$ uniformly for every polynomial $p$. Given $f\in C[0,1]$ and $\epsilon >0$, choose a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. The operator norm estimate in (a) givesFirst let $n\to \infty$, then $\epsilon \downarrow 0$. Hence $\Vert B_{n}f-f{\Vert }_{\infty }\to 0$ for every continuous $f$. This is an instance of extension of approximation operators from a dense subspace.