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A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

A normed vector space is a vector space equipped with a norm, whose induced metric is $d(x,y)=\Vert x-y\Vert$. The norm must be positive on nonzero vectors, scale by the absolute value of a scalar, and satisfy the triangle inequality. These conditions make distance compatible with addition and scaling.

On $C([0,1])$, both the supremum norm $\Vert f{\Vert }_{\infty }={\max }_{0\le x\le 1}|f(x)|$ and the L2 norm $\Vert f{\Vert }_2=({\int }_0^1|f(x){|}^{2}dx{)}^{1/2}$ are genuine norms. A nonzero continuous function stays nonzero on an interval of positive length, so its squared integral cannot vanish. For $f(x)=x$, they give $1$ and $1/\sqrt{3}$, respectively. Always $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$, but the shrinking peaks below show that no uniform reverse bound exists. Unlike the finite-dimensional equivalence of norms, infinite-dimensional vector spaces can support inequivalent norm topologies.

For real continuous $f$, the Chebyshev alternation theorem says that $p\in {\mathcal{P}}_n$ is the unique best uniform approximation to $f$ precisely when there are $n+2$ ordered points $0\le x_0<\cdots <x_{n+1}\le 1$ and a sign $\sigma \in \{-1,1\}$ such thatHere ${\mathcal{P}}_n$ means degree at most $n$. Best approximants exist: a minimizing sequence is bounded in the finite-dimensional polynomial space, and finite-dimensional equivalence of norms gives a convergent coefficient subsequence. When $f=p$, the zero error is treated separately; the derivative-sign hypothesis below excludes that case.

Let $p_n$ be a best degree-at-most-$n$ approximant and suppose for contradiction that $E_n(f)=E_{n+1}(f)$. Since $f^{(n+1)}>0$, $f$ is not a degree-at-most-$n$ polynomial and its error is nonzero. The same $p_n$ would be best in ${\mathcal{P}}_{n+1}$, so the Chebyshev alternation theorem at that degree supplies $n+3$ alternating error extrema. The intermediate value theorem gives at least $n+2$ distinct zeros of $e=f-p_n$, one between each consecutive pair of extrema.

Repeated Rolle's theorem then forces a zero of $e^{(n+1)}$. But $p_n^{(n+1)}=0$, whereas $e^{(n+1)}=f^{(n+1)}>0$ everywhere. This contradiction provesThe argument proves strict decrease of best polynomial approximation under a derivative sign; an everywhere negative $(n+1)$st derivative gives the same strict conclusion after replacing $f$ by $-f$.

Choose a best uniform approximation $p_n\in {\mathcal{P}}_n$. It exists because this vector subspace is a finite-dimensional vector space and a minimizing sequence $q_j$ is bounded: $\Vert q_j{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }+\Vert f-q_j{\Vert }_{\infty }$. Finite-dimensional equivalence of norms then supplies a convergent subsequence of coefficients. If $E_n(f)>0$, the Chebyshev alternation theorem gives $n+2$ alternating error extrema. The intermediate value theorem supplies a zero $x_{n,k}$ of $f-p_n$ in each of the $n+1$ intervening open intervals. These interpolation nodes are distinct, and uniqueness of polynomial interpolation givesThe convergence follows from the Weierstrass approximation theorem. If $E_n(f)=0$, choose any $n+1$ distinct nodes: the polynomial interpolant is still $f=p_n$. This constructs interpolation nodes from best uniform approximation; the nodes depend on $f$, and no universal node system for all continuous functions is asserted.

Two norms $\Vert \cdot {\Vert }_a,\Vert \cdot {\Vert }_b$ are Lipschitz equivalent norms when constants $c,C>0$ satisfy $c\Vert x{\Vert }_a\le \Vert x{\Vert }_b\le C\Vert x{\Vert }_a$ for every vector $x$. Applying these inequalities to $x_j-x_k$ shows that a sequence is a Cauchy sequence for one norm exactly when it is so for the other; applying them to $x_j-x$ likewise equates convergence. If one norm is complete, a Cauchy sequence for the other converges using the first norm and hence also the second. Interchanging the norms proves completeness is preserved in both directions.

Write $x={\sum }_i{x}_i{e}_i$ and set $C={\sum }_i\Vert e_i\Vert$. If $\Vert x{\Vert }_2=1$, then $|x_i|\le 1$, and the triangle inequality givesScaling a nonzero $x$ gives $\Vert x\Vert \le C\Vert x{\Vert }_2$ for every $x$. The reverse triangle inequality now impliesso $x\mapsto \Vert x\Vert$ is continuous, indeed Lipschitz, in the Euclidean norm.

Suppose there were no positive lower bound on the Euclidean unit sphere. We could choose $x_j$ with $\Vert x_j{\Vert }_2=1$ but $\Vert x_j\Vert <1/j$. The Bolzano-Weierstrass theorem supplies a subsequence converging in the Euclidean norm to $x$. Continuity of both norms gives $\Vert x{\Vert }_2=1$ and $\Vert x\Vert =0$, contradicting definiteness. Therefore some $m>0$ satisfies $\Vert x\Vert \ge m$ on the unit sphere. Scaling again proves the finite-dimensional equivalence of norms:

Let $P_n$ be the finite-dimensional vector space of real polynomials of degree at most $n$. For fixed $a>0$,is a norm: if it vanishes, the polynomial vanishes on an interval and hence is the zero polynomial. Alsois a norm on $P_n$. By finite-dimensional equivalence of norms, there is $C>0$, depending only on $n$ and $a$, such that $M(p)\le CN(p)$. ThereforeThe extreme value theorem supplies $y\in [0,a]$ at which the last supremum is attained, and hence