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The continuous dual space $X^{\ast }$ consists of all bounded real linear functionals on $X$, with operator norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. Let $(f_n)$ be Cauchy in this norm. For each $x$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $f_n(x)$ converges in $\mathbb{R}$; define its limit to be $f(x)$.

Pointwise limits of the linear identities show that $f$ is linear. The sequence of norms is bounded, say by $M$, so $|f(x)|\le M\Vert x\Vert$ and $f\in X^{\ast }$. Given $\epsilon >0$, choose $N$ so that $\Vert f_n-f_m\Vert \le \epsilon$ for all $m,n\ge N$. Holding $n$ fixed and taking the pointwise limit in $m$ gives $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ for all $x$. Thus $\Vert f_n-f\Vert \le \epsilon$, provingCompleteness of $X$ is not needed; completeness of the scalar field supplies the pointwise limit.

For a functional $E$ on a real Banach space, its convex conjugate is defined on the dual space byHere $⟨p,u⟩$ is the dual pairing. For complex spaces, use its real part, equivalently regard the space as real for convex analysis. The convex conjugate is a convex function, being the supremum of affine functions of $p$, even when $E$ itself is not convex.

Write ${\iota }_C$ for the indicator functional of a constraint set $C$ and $q(u)=\Vert u{\Vert }^2/2$. ThenBoth terms are convex, hence their infimal convolution is the convex squared distance to a convex set. The closest point theorem in a Hilbert space supplies the unique metric projection onto a closed convex set $P_{C}u$ and gives $d_C(u)=\Vert u-P_{C}u{\Vert }^2/2$.

Identify the Hilbert space with its dual space using the Riesz representation theorem. Completing the square yields $q^{\ast }(p)=\Vert p{\Vert }^2/2$, while ${\iota }_C^{\ast }(p)={\sup }_{v\in C}⟨p,v⟩={\sigma }_C(p)$ is the support function. The conjugate of the squared distance to a convex set is thereforeFor the closed unit ball, the metric projection onto a closed convex set is $P_{C}u=u$ if $\Vert u\Vert \le 1$, and $P_{C}u=u/\Vert u\Vert$ otherwise. ThereforeThe last equality uses the Cauchy-Schwarz inequality to compute the unit ball's support function, attained in the direction of $p$ when $p\ne 0$.

Choose a countable dense subset $\{x_1,x_2,\dots \}$ of the normed vector space $X$. For each fixed $j$, the scalar sequence $(f_n(x_j))$ is bounded becauseRepeated use of the Bolzano-Weierstrass theorem followed by a diagonal subsequence argument produces a subsequence $(f_n{)}_{n\in \Lambda }$ for which $f_n(x_j)$ converges for every $j$.

Let $D$ be the linear span of the $x_j$. On every $x\in D$, defineFinite linear combinations show that $f:D\to \mathbb{R}$ is linear, and passage to the limit in $|f_n(x)|\le \Vert x\Vert$ gives $|f(x)|\le \Vert x\Vert$. Thus $f$ is a bounded linear functional on the dense linear subspace $D$ and extends uniquely to an element of the dual space $X^{\ast }$ with $\Vert f\Vert \le 1$.

For arbitrary $x\in X$ and $y\in D$,First choose $y$ close to $x$, then let $n\in \Lambda$ tend to infinity. This provesThis is the Sequential Banach-Alaoglu theorem for a separable predual.

For $u\in U$, define $\varphi (u)\in U^{\ast }$ by $\varphi (u)(u^{\prime })=⟨u,u^{\prime }⟩$. If $\varphi (u)=0$, then in particularso $u=0$. Thus $\varphi$ is injective. Since a finite-dimensional vector space and its dual space have the same dimension, $\varphi$ is also surjective and hence an isomorphism. This is the finite-dimensional real case of the Riesz representation theorem.

The adjoint operator of $\alpha :V\to W$ is the unique linear map ${\alpha }^{\ast }:W\to V$ satisfyingfor every $v\in V$ and $w\in W$. In the stated orthonormal bases, if $v$ and $w$ denote coordinate columns, thenTherefore the matrix of ${\alpha }^{\ast }$ is the matrix transpose $A^T$.

For $w\in W$,Hence $\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\perp }$. Taking orthogonal complements in the finite-dimensional space $W$ proves the image-kernel orthogonality for an adjoint

Put $r=\alpha (v_0)-w_0$. For any $h\in V$,This is minimized at $h=0$ exactly when $r\perp \mathrm{im}\alpha$, equivalently when ${\alpha }^{\ast }r=0$. Thus the normal equation for a linear inverse problem is

For the given linear least-squares problem,The normal equations $2x+y=3$ and $x+2y=5$ have the unique solution

A bilinear form $B$ on $V$ is nondegenerate whenIn finite dimensions this is equivalent to nondegeneracy in the first argument.

Define the linear mapNondegeneracy makes ${\beta }_1$ injective. Since $V$ and its dual space have the same finite dimension, ${\beta }_1$ is an isomorphism. Similarly define ${\beta }_2(w)=B_2(-,w)$ and setThen $\alpha$ is linear and

If $w\in \ker \alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. HenceThis is the representation of a bilinear form relative to a nondegenerate bilinear form.

For the log-cosh energy on a Sobolev space, the elementary inequalities $0\le \log \cosh t\le t^2/2$ follow by integrating $|\tanh t|\le |t|$ from zero. Thusso the energy is well-defined on $H^1$. For a fixed $v$, its proposed derivative is linear in the direction $h$ and bounded, because $|\tanh v|\le |v|$ givesTo justify differentiation, use $(\log \cosh t{)}^{\prime \prime }={\mathrm{sech}}^{2}t\in (0,1]$. Taylor's integral remainder pointwise, combined with the exact quadratic gradient expansion, givesEvery term is integrable by the displayed bounds. The remainder is quadratic in the norm, so the functional is even Fréchet differentiable, and in particular Gâteaux differentiable with the required derivative. This also proves that $A(v)$ belongs to the stated dual space.

For $v,w\in H^1$,Both terms are nonnegative, and the integrand in the second is strictly positive wherever $v\ne w$, since $\tanh$ is strictly increasing. Distinct Sobolev elements differ on a set of positive measure, so the expression is positive. Hence $A$ is strictly monotone.

This gives strict monotonicity without coercivity on H1: it is not coercive on $H^1$. For constant functions $v_t=t$ with $t\to +\infty$, the norm is $t|\Omega {|}^{1/2}$ andwhich is finite. In contrast it is coercive on $H_0^1$: if $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, thenDividing by the norm proves growth to infinity. On this smaller space the operator is understood by restricting its dual action to zero-trace test functions. Removing the constant directions by the boundary condition is what restores coercivity.

For finite-dimensional vector spaces $V,W$ over a field $F$, a bilinear pairing $B:V\times W\to F$ is perfect when the induced linear maps $V\to W^{\ast }$ and $W\to V^{\ast }$ are isomorphisms. Equivalently, both dimensions agree and the matrix of the pairing in any two bases is invertible. The evaluation pairing between a finite-dimensional vector space and its dual space is perfect. This notion allows two different spaces; a nondegenerate bilinear form is the case $V=W$.

For a closed oriented $d$-dimensional manifold over a field $F$, the cup product pairingis a perfect pairing between complementary degrees. The cap product with the fundamental class is an isomorphism by Poincare duality, and the universal coefficient theorem for cohomology over $F$ identifies the complementary cohomology with the full dual space of the resulting homology. Evaluation is therefore a perfect pairing. Taking the top-degree part of the same formula defines a nondegenerate bilinear form on the full graded cohomology. A boundary requires relative groups and Poincare-Lefschetz duality instead.

A linear subspace $W\subseteq X^{\ast }$ is separating if every nonzero $x\in X$ has $T(x)\ne 0$ for some $T\in W$. For finite-dimensional $X$, an evaluation map $X\to F^{\dim W}$ is injective, so the rank-nullity theorem forces $\dim W\ge \dim X=\dim X^{\ast }$ and $W=X^{\ast }$. For the infinite-dimensional polynomial space, the span of its coefficient functionals is a proper separating linear subspace of the dual space, which contains all sequences of coefficients, not just finitely supported ones.

The continuous dual space of $H^s({\mathbb{R}}^d)$ is $H^{-s}({\mathbb{R}}^d)$ under the extension of the L2 inner product. The Fourier transform definition $\Vert f{\Vert }_{H^s}^2=\int (1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d\xi$ proves boundedness of the pairing by the Cauchy-Schwarz inequality, and the Riesz representation theorem proves that every continuous functional has this form. Consequently a bounded linear operator $K:H^r\to H^{r+a}$ has an adjoint operator, relative to this pairing, from $H^{-r-a}$ to $H^{-r}$. For an operator that smooths by two orders for every $r$, this yields $K^{\ast }:H^2\to H^4$ by taking $r=-4$.

For a finite-dimensional symplectic vector space, the symplectic adjoint of a linear map $T$ is the unique linear map $T^{\ast }$ satisfying the displayed identity. Existence and uniqueness follow because the nondegenerate bilinear form identifies the vector space with its dual space. If $\omega$ has matrix $J$ in a basis, then $T^{\ast }=J^{-1}{T}^{\mathsf{T}}J$. Thus $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$ and $(T^{\ast }{)}^{\ast }=T$, and membership in the symplectic Lie algebra is equivalent to $T^{\ast }=-T$. This operation is linear over the ground field; it does not involve complex conjugation. It preserves diagonalizability and the property of being a nilpotent linear map, since it is conjugate to matrix transpose.

A linear map on a finite-dimensional vector space over an algebraically closed field has both an upper triangular matrix and a lower triangular matrix in suitable bases. One proof uses invariant subspaces and annihilators to make the induction step entirely in the dual space.

Choose an eigenvector $v$ and let $U=\mathrm{span}\{v\}$. The dual map preserves the annihilator of a vector subspace $U^0$, of dimension one less. By induction, choose a basis of $U^0$ making the restriction upper triangular. Extend this to a basis of the whole dual space; its matrix $C$ remains upper triangular because $U^0$ is an invariant subspace. In the corresponding primal basis, the matrix is $C^T$, which is lower triangular. Reversing the primal basis makes it upper triangular. Dimension zero and one start the induction. This also allows the original eigenvector to be the first vector of the resulting upper triangular basis.