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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.

For a L2-bounded continuous martingale, the terminal and maximal norms satisfyThe first inequality follows from almost sure convergence. Apply the Doob L2 maximal inequality on $[0,T]$ and then the monotone convergence theorem as $T\to \infty$ for the second. These are equivalent norms on the space modulo indistinguishability of stochastic processes; neither inequality needs a zero initial value.

For every finite $d\ge 1$, a subset $K\subseteq {\mathbb{R}}^d$ of Euclidean space is compact if and only if it is a closed set and a bounded set. Here closed means closed in the ambient Euclidean space, not merely in $K$ itself, and the bounded set condition uses the Euclidean norm. The empty set is included: it is a closed set, a bounded set, and compact. ^[ref]

This makes an apparently infinite test practical. To verify compactness of a closed interval $[a,b]$ with finite endpoints, we need not examine every open cover: it is closed in the real line and a bounded set. By contrast, the open interval $(0,1)$ is a bounded set but is not closed in the real line, as the missing-endpoint figure in compact space illustrates. The whole real line is closed but is not a bounded set, so it too fails compactness. These examples show why both hypotheses are needed.

The same criterion holds in any real or complex finite-dimensional vector space equipped with a norm: all norms there are equivalent norms, so they give the same open sets, closed sets, and bounded sets. Since compactness depends only on the topology, choosing a different norm does not change the conclusion. Without an appropriate topology, a bare vector space alone has no notion of compactness.

The restriction to a finite-dimensional vector space is essential. In the l2 sequence space, an infinite-dimensional vector space that is also a Hilbert space, let $e_n$ be the vectors of its standard orthonormal basis. All lie in the closed unit ball $B=\{x:\Vert x\Vert \le 1\}$, which is a closed set and a bounded set in the norm topology. For distinct indices, their inner product is zero, givingThus distinct terms remain distance $\sqrt{2}$ apart. No subsequence is a Cauchy sequence, hence none is a convergent sequence. The equivalence of compactness and sequential compactness in a metric space proves that this closed unit ball is not compact. The conclusion concerns the norm topology.

A short proof outline in Euclidean space explains the role of a finite-dimensional vector space. For a sequence in a bounded set, apply the Bolzano-Weierstrass theorem to each of the Cartesian coordinates in turn, retaining a subsequence at each of the finitely many steps. The final subsequence is a convergent sequence in all Cartesian coordinates, and a closed set contains its limit of a sequence. This yields sequential compactness, hence compactness. Conversely, a compact subset of a metric space is a closed set and a bounded set. ^[ref]

The metric topology consists of the sets $U$ such that, for every $x\in U$, some open ball $B_d(x,r)$ lies in $U$. Thus open sets describe which sufficiently small changes remain inside a region; they do not specify the numerical size of those changes.

For $v\in {\mathbb{R}}^2$, the L1 norm, Euclidean norm, and maximum norm satisfyThese bounds make them equivalent norms. For example, $B_{\infty }(a,r/2)\subseteq B_1(a,r)\subseteq B_{\infty }(a,r)$: every open ball contains a sufficiently small open ball for either other metric. Hence all three define the same open sets, convergent sequences, and continuous maps. More generally, any two norms on a real or complex finite-dimensional vector space are equivalent norms; see H. A. Priestley, Functional Analysis I, §§0.5–0.7 and Theorem 4.3.

The metric topology therefore does not determine numerical distances. Multiplying a metric by $100$ preserves its open sets while multiplying every distance by $100$. Nor does the general equivalence of metrics require constant-factor bounds: that stronger comparison holds for the three norms used here.

By equivalent norms, choose $c,C>0$ with $c\Vert h{\Vert }_1\le \Vert h{\Vert }_2\le C\Vert h{\Vert }_1$. Differentiability in the first norm means that there is a bounded linear map $L:V\to \mathbb{R}$ with $f(a+h)-f(a)-Lh=r(h)$ and $|r(h)|/\Vert h{\Vert }_1\to 0$. The same $L$ is bounded in the second norm, since $|Lh|\le K\Vert h{\Vert }_1\le K{c}^{-1}\Vert h{\Vert }_2$. Moreover,The two norms define the same approach $h\to 0$, so this proves differentiability with the same Fréchet derivative in the second norm. Reversing the norms proves the converse.

For completeness, the derivative is unique: if $L_1,L_2$ both work, fix $v$ and take $h=tv$. The two remainder estimates imply $|(L_1-L_2)v|=o(1)$ as $t\to 0$, so $(L_1-L_2)v=0$ for every $v$. Thus differentiability and its derivative are independent of the chosen norm. This is differentiability under equivalent norms, not merely invariance of continuity.

Adjoin the identity to the bounded multiplicatively closed set, putting $S^{\prime }=S\cup \{1\}$ and $M={\sup }_{s\in S^{\prime }}\Vert s\Vert <\infty$. Define an auxiliary norm on the underlying Banach space byThe identity ensures $\Vert a\Vert \le p(a)\le M\Vert a\Vert$, so $p$ and the given norm are equivalent norms. For $s\in S$, closure under multiplication gives $p(sa)\le p(a)$. Now use the operator norm of left multiplication on this renormed space:Left multiplication is faithful, since $a1=a$. It follows that this is a norm, its triangle and scalar laws follow from the operator norm, and $L_{ab}=L_a{L}_b$ proves submultiplicativity. Also $L_1$ is the identity, so $\Vert 1{\Vert }_1=1$, and the contraction property gives $\Vert s{\Vert }_1\le 1$ for every $s\in S$. Finally,For the upper bound use $p(ab)\le M\Vert a\Vert \Vert b\Vert \le M\Vert a\Vert p(b)$. These bounds prove equivalence and completeness, so this is the required unital algebra norm. This proves renorming a bounded multiplicative semigroup; the auxiliary norm is submultiplicative but need not give the identity norm one, which is why the operator norm is used.

The spectrum of an element is $\sigma (x)=\{\lambda \in \mathbb{C}:\lambda 1-x\text{ is not invertible in }A\}$. The spectral radius formula isIf $r(x)<1$, choose $r(x)<q<1$. Eventually $\Vert x^m{\Vert }^{1/m}<q$, hence $\Vert x^m\Vert <q^m$. The finitely many earlier powers are bounded too. In particular the power set is bounded; indeed $x^m\to 0$ in norm.

For the commuting family, put $d_j=r(x_j)+\epsilon /2$ and $y_j=x_j/d_j$. Each $r(y_j)<1$, so $M_j={\sup }_{m\ge 0}\Vert y_j^m\Vert <\infty$. Every element of the generated multiplicative semigroup can be reordered as $y_1^{m_1}\cdots y_n^{m_n}$, and its norm is at most ${\prod }_j{M}_j$. The first construction therefore gives one equivalent unital algebra norm with $\Vert y_j{\Vert }_0\le 1$ for all $j$, whenceThis is simultaneous spectral-radius renorming; commutativity is what bounds all mixed products.

For commuting $a,b$, apply this construction to both with error $\epsilon$. Equivalent algebra norms leave invertibility and spectrum unchanged, and the spectral radius formula gives $r(c)\le \Vert c{\Vert }_0$. ThusLet $\epsilon \downarrow 0$ to obtain $r(ab)\le r(a)r(b)$ and $r(a+b)\le r(a)+r(b)$.

The norm comparison in bounded distortion of a Banach space is understood up to a positive rescaling of the new norm. Thus, after replacing $D$ by $D^2$ if one's convention uses symmetric comparison constants, we may use the following form: for every equivalent norm $q$ on a block subspace, a further block subspace $Z$ and a number $a>0$ satisfyThe number $a$ may depend on the norm and the subspace; $D$ is fixed. Without this rescaling convention the premise would be impossible, since arbitrarily large scalar multiples of the original norm are equivalent norms. We prove the substantive, scale-invariant assertion for an infinite Schauder basis.

Let $K$ be the basis constant of the original basis. Every block sequence has basis constant at most $K$, so its initial, tail and finite interval projections have norms at most $K$, $1+K$ and $2K$, respectively. We shall construct increasingly complicated finite unconditional configurations while keeping their unconditional constant fixed at $C=2D$.

First construct the Maurey hierarchy of finite sets. Write ${\mathcal{M}}_0=\{\varnothing \}$, and definewhere prepending to the empty set is permitted. For each nonzero countable limit ordinal $\lambda$, fix positive ordinals ${\lambda }_j\uparrow \lambda$ and putInduction shows that every family is hereditary and spreading: subsets remain in the family, and increasing any of the coordinates preserves membership. Every nonzero stage contains all singletons. These families are also well-founded as extension trees. At a successor stage, an infinite branch would yield an infinite branch at the previous stage after deleting its first integer. At a limit stage, a hypothetical branch with first integer $m$ has every initial segment in the finite union ${\bigcup }_{j\le m}{\mathcal{M}}_{{\lambda }_j}$. One family then contains arbitrarily long initial segments, and heredity would give it an infinite branch.

For every infinite set $L$, the extension tree ${\mathcal{M}}_{\alpha }\cap [L{]}^{<\infty }$ has empty-root rank of a well-founded tree at least $\alpha$. Here is the induction proving this lower bound. At a successor stage fix $n\in L$; the tree above the node $\{n\}$ contains a copy of ${\mathcal{M}}_{\alpha }$ restricted to the infinite tail of $L$ after $n$, and hence has rank at least $\alpha$. The root therefore has rank at least $\alpha +1$. At a limit stage, for every $j$ the restriction of ${\mathcal{M}}_{{\lambda }_j}$ to a sufficiently late infinite tail of $L$ is contained in ${\mathcal{M}}_{\lambda }$. The root rank is consequently at least ${\sup }_j{\lambda }_j=\lambda$.

The crucial elementary estimate concerns signed interval projection norms. Given a hereditary family $\mathcal{F}$ containing all singletons and a basis $(w_i)$, define on finitely supported vectorsChoosing one interval containing the support gives $q_{\mathcal{F}}^{(w)}(x)\ge \Vert x\Vert$. Whenever this supremum is bounded by a multiple of $\Vert x\Vert$, it extends to a norm on the closed span giving equivalent norms with the original norm.

If $S={\sum }_i{\epsilon }_i{P}_{E_i}$ is one of the admitted signed interval projections, thenTo prove this, test $Sx$ with another admitted operator $R={\sum }_j{\delta }_j{P}_{F_j}$. The composition is the signed sum of projections onto the nonempty intervals $E_i\cap F_j$. Colour each intersection according to which of $E_i,F_j$ supplies its left endpoint, breaking ties in favour of $E_i$. In the first colour the list of minima is a subset of $(\min E_i)$, with no repetitions; in the second it is a subset of $(\min F_j)$. Both lists belong to $\mathcal{F}$ by heredity. Each colour therefore contributes an admitted signed interval projection, and the triangle inequality gives $\Vert RSx\Vert \le 2q_{\mathcal{F}}^{(w)}(x)$. Taking the supremum over $R$ proves the estimate.

Now suppose $q_{\mathcal{F}}^{(w)}$ is an equivalent norm on a block subspace $W$. Apply bounded distortion to obtain a further block basis $(z_i)$ spanning $Z$, with $a\Vert z\Vert \le q_{\mathcal{F}}^{(w)}(z)\le aD\Vert z\Vert$. A signed interval projection relative to $(z_i)$, with an admissible list of minima, is the restriction of an admitted signed interval projection relative to $(w_i)$. Indeed, replace each interval of block indices by the interval from the first to the last original coordinate of its blocks; its minimum only increases, so the spreading property preserves admissibility. The elementary estimate and the norm comparison therefore giveIn particular $q_{\mathcal{F}}^{(z)}\le C\Vert \cdot \Vert$, with the same $C=2D$ for every family to which we apply this procedure.

We prove by transfinite induction that, in every block subspace and for every $1\le \alpha <{\omega }_1$, there is a normalized block basis $(z_i)$ for whichFor $\alpha =1$ only one interval is permitted, so the initial signed interval projection norm is bounded by $2K\Vert x\Vert$. Apply the preceding renorming procedure.

For a successor stage, first choose $(w_i)$ satisfying the induction hypothesis for ${\mathcal{M}}_{\alpha }$. Delete the first interval from any ${\mathcal{M}}_{\alpha +1}$-admissible family. The remaining minima lie in ${\mathcal{M}}_{\alpha }$, soThis is an equivalent norm. The renorming procedure produces a further block basis with the constant reset to $C$.

At a limit stage $\lambda$, choose decreasing block subspaces $W_j$, each with a block basis satisfying the induction hypothesis for ${\mathcal{M}}_{{\lambda }_j}$. Choose successive normalized $z_j\in W_j$. Make each $z_j$ sufficiently far out that its first coordinate in the basis of every $W_k$, $k\le j$, is at least $j$; only finitely many restrictions are imposed at each choice. An admissible interval family for ${\mathcal{M}}_{\lambda }$ belongs to ${\mathcal{M}}_{{\lambda }_j}$ for some $j\le \min E_1$. All blocks on which it acts lie in the tail $[z_i:i\ge j]\subset W_j$. Transferring its intervals to the basis of $W_j$ increases their minima, so the induction hypothesis givesThus $q_{{\mathcal{M}}_{\lambda }}^{(z)}$ is again an equivalent norm, and one more application of bounded distortion resets its bound to $C$. This completes the transfinite induction. The tail-projection factor in this step is important; the diagonal tail need not have the same norm as the original vector.

Finally let $T_C(X)$ be the tree of finite normalized sequences $(x_1,\dots ,x_n)$ satisfyingfor all scalars and all signs. Each finite level is closed in the corresponding power of the unit sphere, since the norm inequalities are closed conditions. The unit sphere of the separable Banach space $X$ is a Polish space. An infinite branch is an unconditional basic sequence: sign changes give coordinate projections with norm at most $(1+C)/2$, imply linear independence, and give uniformly bounded initial projections, hence the basis property on the closed span. These projection bounds also give unconditional convergence, including over complex scalars.

For each countable $\alpha$, the block basis constructed above maps the extension tree ${\mathcal{M}}_{\alpha }$ into $T_C(X)$ by $\{i_1<\cdots <i_r\}\mapsto (z_{i_1},\dots ,z_{i_r})$. To check the sign inequality, use the admissible singleton intervals $\{i_1\},\dots ,\{i_r\}$ on the vector $\sum a_k{z}_{i_k}$. Thus the unconditional tree index is at least $\alpha$ for every countable ordinal. If $T_C(X)$ were well-founded, the closed-tree result would give it countable height, contradicting these lower bounds. It has an infinite branch. Therefore $X$ contains an unconditional basic sequence.

Two norms $N_1,N_2$ on the same vector space are equivalent norms, or Lipschitz equivalent, if there are constants $0<c\le C<\infty$ such that $c{N}_1(v)\le N_2(v)\le C{N}_1(v)$ for every $v$. The constants are uniform over the whole space.

Let $N$ be any norm on ${\mathbb{R}}^n$. The triangle inequality and the Cauchy-Schwarz inequality giveThe reverse triangle inequality then gives $|N(x)-N(y)|\le C\Vert x-y{\Vert }_2$, so $N$ is Euclidean-continuous; this continuity was derived, rather than assumed. On the compact Euclidean unit sphere, $N$ attains a strictly positive minimum $c$, because it vanishes only at zero. Homogeneity yieldsFinally a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ satisfies $\Vert Ax{\Vert }_2\le ({\sum }_{i,j}|a_{ij}{|}^2{)}^{1/2}\Vert x{\Vert }_2$, again by Cauchy-Schwarz inequality. The equivalence just proved transfers this bound to any norms on the source and target. Hence every such linear map is continuous.

Write $N_2=\Vert M_{\infty }{\Vert }_2$ and $N_{\ast }=\Vert {\sup }_{t\ge 0}|M_t|{\Vert }_2$. The L2 martingale convergence theorem ensures the existence of $M_{\infty }$ and yields $|M_{\infty }|\le {\sup }_t|M_t|$ almost surely. Hence $N_2\le N_{\ast }$.

For a finite horizon $T$, the Doob L2 maximal inequality for the submartingale $|M|$ givesThe inequality does not require $M_0=0$. Send $T$ to infinity: the left side increases to $N_{\ast }^2$ by the monotone convergence theorem, and the right side tends to $4N_2^2$ by convergence in the Lebesgue space $L^2$. Thus the equivalent terminal and maximal norms for L2-bounded continuous martingales satisfyBoth are genuine norms on L2-bounded continuous martingales modulo indistinguishability of stochastic processes: a zero terminal norm gives $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]=0$, first at rational times and then at all times by continuity. This proves that the two norms are equivalent norms.

Two norms $N_1,N_2$ are equivalent norms when constants $c,C>0$ satisfyfor every $v$.

For $x,y\in {\mathbb{R}}^n$, the reverse triangle inequality givesby the Cauchy-Schwarz inequality. Hence $g$ is Lipschitz continuous, and therefore continuous.

On the compact Euclidean unit sphere, $g$ is positive and continuous, so the extreme value theorem gives $0<m\le g\le M<\infty$. Homogeneity then yieldsThus every norm on a finite-dimensional vector space is equivalent to the Euclidean norm; comparing two such bounds proves any two norms on $V$ are Lipschitz equivalent.

Two norms $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$ on one vector space are Lipschitz equivalent when constants $c,C>0$ exist such that

For $1\le p<q<\infty$, the same finitely supported sequence givesOne of the two inequalities required for equivalent norms therefore fails. Hence

Since $1\le t\le b$ and $c\ge 0$,Taking suprema givesThus the weighted norm and the usual uniform norm are equivalent norms.

For $x\in X$, define $e_x\in X^{\ast \ast }$ by $e_x(\varphi )=\varphi (x)$. The map $J:x\mapsto e_x$ is linear, andshows $\Vert Jx\Vert \le \Vert x\Vert$, so $J$ is continuous. The granted equality $\Vert Jx\Vert =\Vert x\Vert$ is the canonical embedding into the bidual.

Suppose, contrary to the claim, that exactly the same linear functionals are continuous for $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$. Their common continuous dual has two operator norms, say $\Vert \cdot {\Vert }_{\ast }$ and $\Vert \cdot {\Vert }_{\ast }^{\prime }$. Both dual spaces are Banach, even if the second normed space $X$ is not complete. The identity maphas a closed graph: convergence in either operator norm implies pointwise convergence on $X$, so two limits must agree. The closed graph theorem makes the two dual norms equivalent.

Applying the isometric bidual formula for each primal norm then gives constants $c,C>0$ such thatThis says the two norms are equivalent norms, contrary to the hypothesis. Their continuous duals must therefore differ as sets, so a linear functional belonging to one and not the other is continuous for exactly one of the two norms.