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The block mixed norm sums the ${\ell }^q$ norm of each block, for $q\ge 1$. Its dual norm is $\Vert p{\Vert }_{q^{\prime },\infty }={\max }_i\Vert p_i{\Vert }_{q^{\prime }}$, where $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Apply Holder inequality blockwise and choose a norming block to attain equality. Absolute values are essential for signed components.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

A dual norm measures how strongly a bounded linear functional can respond to a unit-sized input. If $p$ is a bounded linear functional on a normed vector space $X$, write $⟨p,x⟩=p(x)$ and defineIt is the operator norm of $p$ when its scalar output is measured by absolute value. On ${\mathbb{R}}^n$, represent $p(x)=a^{T}x$ using a coefficient vector $a$. The unit ball is symmetric, so maximizing $|a^{T}x|$ gives the same value as maximizing $a^{T}x$: negate an input whenever its output is negative.^[ref]

For an original two-coordinate example, fix $a=(2,-1{)}^T$, so $p(x)=2x_1-x_2$. The input norm changes the allowable unit ball and hence the dual norm:Under the L1 norm, $|2x_1-x_2|\le 2(|x_1|+|x_2|)\le 2$, and $(1,0{)}^T$ attains the bound. Under the Euclidean norm, the Cauchy-Schwarz inequality gives $|a^{T}x|\le \Vert a{\Vert }_2\Vert x{\Vert }_2\le \sqrt{5}$, with equality at the displayed normalized vector. Under the maximum norm, each coordinate can reach unit absolute value, so $|2x_1-x_2|\le 2+1=3$; $(1,-1{)}^T$ attains it. Thus the dual norms of these three vector norms are respectively the maximum norm, the Euclidean norm, and the L1 norm. For $1\le p\le \infty$, Lp norms on a finite coordinate space pair with their Holder conjugate exponents.^[ref]

This geometry explains the usefulness of a dual norm: it translates a bound on inputs into the sharp bound on a linear output. The support function of the unit ball gives its value in the coefficient direction $a$. Changing the input budget changes the set of possible inputs; the three different answers do not conflict. In finite dimensions a closed unit ball is compact, so a continuous linear functional achieves its maximum absolute value. In an infinite-dimensional normed vector space, the supremum remains the definition and need not be achieved.

Suppose a perturbation $\delta x$ satisfies $\Vert \delta x\Vert \le \epsilon$, with $\epsilon \ge 0$. Absolute homogeneity of a norm and the dual norm definition giveFor nonzero $\delta x$, divide it by its norm to obtain an input in the unit ball; for zero perturbation the inequality is immediate. This is an exact error bound for a linear functional, for every perturbation in the stated input set.

For $p(x)=2x_1-x_2$ and a budget $\epsilon =0.1$, the sharp output-error bounds are $0.2$ for $\Vert \delta x{\Vert }_1\le 0.1$, $\sqrt{5}/10\simeq 0.224$ for $\Vert \delta x{\Vert }_2\le 0.1$, and $0.3$ for $\Vert \delta x{\Vert }_{\infty }\le 0.1$. Scale the maximizing vectors from the preceding example by $0.1$ to attain the positive endpoints; their negatives attain the negative endpoints. The L1 norm budget constrains a total absolute error; the maximum norm budget permits each coordinate to have that error separately. These budgets define nested sets and carry different information.

The coordinates must use the stated scale and units. Rescaling one coordinate changes both its coefficient in $a$ and the meaning of its input budget. For a nonlinear differentiable function with scalar values, its Fréchet derivative at the base point is a linear functional governing the first-order response. An exact finite-perturbation error bound additionally needs control of the Taylor remainder or of the Fréchet derivative along the path.

We prove existence by the direct method in the calculus of variations, making the step with weakly compact sets explicit. For $X=W_0^{1,4}(\Omega )$ and $f\in L^{8/7}(\Omega )$, introduce the functionalIt is finite by part (b)(i). Hölder's inequality and the Sobolev embedding theorem giveThus $E$ is a coercive function, bounded below, and every minimizing sequence is bounded in $X$.

The space $X$ is a reflexive Banach space: it is isomorphic to a closed linear subspace of the finite product of the reflexive spaces $L^4(\Omega )$ recording the function and its weak derivatives. A bounded sequence in a reflexive Banach space has a weakly convergent subsequence. Hence, after passing to a subsequence, a minimizing sequence satisfies $u_j\rightharpoonup u_{\ast }$ in $X$. The continuous linear maps $u\mapsto \nabla u$ into $L^4$ and $u\mapsto u$ into $L^8$ preserve weak convergence, soThe norm in any Banach space has weak lower semicontinuity, since it is the supremum of absolute values of bounded linear functionals of dual norm at most one, by the Hahn-Banach theorem. Therefore the two nonnegative norm powers in $E$ have weak lower semicontinuity. Also $u\mapsto {\int }_{\Omega }fudx$ belongs to $X^{\ast }$ and is continuous for weak convergence. It follows thatFor any $v\in X$, the real function $t\mapsto E(u_{\ast }+tv)$ has a minimum at $t=0$. Its derivative there isThis differentiation is justified directly by finite polynomial expansions: the fourth power of the Euclidean norm of $\nabla u_{\ast }+t\nabla v$ and the eighth power of $u_{\ast }+tv$ have integrable coefficients by Hölder's inequality. Thus $A(u_{\ast })=0$, giving existence of a weak solution.

If $u,w$ are two weak solutions, testing the difference by $u-w$ and using part (b)(iii) givesHenceNo sign assumption on $f$ or on the weak solution is required.

For maximization of $f(x)$ subject to $g_j(x)\le 0$ and $h_k(x)=0$, the Lagrangian sufficiency theorem says: if a feasible $x^{\ast }$ globally maximizes $L(x)=f(x)-{\sum }_j{\lambda }_j{g}_j(x)+{\sum }_k{\nu }_k{h}_k(x)$, where ${\lambda }_j\ge 0$, and ${\lambda }_j{g}_j(x^{\ast })=0$, then $x^{\ast }$ globally maximizes $f$ over the feasible set. Indeed, at every feasible $x$,This requires a global maximum of the optimization Lagrangian; a stationary point alone is not sufficient without additional concavity or another justification.

Put $q=p/(p-1)$ and $C=({\sum }_i|a_i{|}^q{)}^{1/q}$. If every $a_i=0$, the maximum is zero and every feasible vector is optimal. Otherwise setSince $p(q-1)=q$, ${\sum }_i|x_i^{\ast }{|}^p=1$. For $g(x)={\sum }_i|x_i{|}^p-1$, the optimization Lagrangian is $L(x)={\sum }_i{a}_i{x}_i-\lambda g(x)$. Its gradient vanishes at $x^{\ast }$ becauseThe function ${\sum }_i|x_i{|}^p$ is a strictly convex function for $p>1$, so $L$ is a strictly concave function. Its supporting-hyperplane inequality proves that this stationary point is its global maximum. Complementary slackness and feasibility hold, so the sufficiency theorem applies. Substitution givesThe maximizing vector is unique when $a\ne 0$. It is the norming vector for Holder inequality, and the value is the corresponding dual norm.

For any vectors $a,b$,Thus, among directions with $\Vert d{\Vert }_{\infty }\le 1$, $g^{T}d\ge -\Vert g{\Vert }_1$. Equality is achieved byFor $g\ne 0$ these minimizing directions have infinity norm one. The strongest decrease per unit infinity-norm step is $-\Vert g{\Vert }_1$, illustrating the dual norm pairing between the infinity norm and the one-norm.

To match the unnormalized convention in part (a), minimize $g^{T}s+\frac{1}{2}\Vert s{\Vert }_{\infty }^2$. At fixed $t=\Vert s{\Vert }_{\infty }$, its minimum is $-t\Vert g{\Vert }_1+\frac{1}{2}{t}^2$, attained at $t=\Vert g{\Vert }_1$. One conventional choice is therefore $\boxed{s=-\Vert g{\Vert }_1\mathrm{sgn}(g)}$, with zero entries where $g_j=0$. Its ray agrees with the unit direction just computed.