Bounded bilinear form ($|B(u,v)|\le C\Vert u\Vert \Vert v\Vert$)

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.

For a finite real scalar matrix $(a_{ij})$ and unit vectors $u_i,v_j$ in a real Hilbert space, $|\sum a_{ij}⟨u_i,v_j⟩|\le K_G{\max }_{{\epsilon }_i,{\delta }_j\in \{-1,1\}}|\sum a_{ij}{\epsilon }_i{\delta }_j|$, with a universal constant. One valid bound is $K_G\le \pi /(2\log (1+\sqrt{2}))$. Expand $\sin (ct)$ using odd tensor powers with coefficient magnitudes summing to $\sinh c=1$, then apply the Gaussian sign identity $\mathbb{E}[\mathrm{sgn}⟨G,u⟩\mathrm{sgn}⟨G,v⟩]=(2/\pi )\arcsin ⟨u,v⟩$. Taking $c=\log (1+\sqrt{2})$ converts the expectation back to the original inner product. Complex scalars admit a universal bound twice this real bound by separating real and imaginary parts.

Every bounded linear operator $T:L^1(\mu )\to H$ into a Hilbert space is an absolutely summing operator, with ${\pi }_1(T)\le K_G\Vert T\Vert$. For simple functions on a common disjoint partition, apply the Grothendieck inequality to the vectors $T(1_A/\mu (A))$ and norming vectors of the images. The scalar sign or phase supremum is the weak 1-norm of the family. Approximation by simple functions proves the general result.

If the closed span of the Rademacher functions were a complemented subspace of $L^1(0,1)$, compose a projection with the inverse of its l2 sequence space parametrization. The Grothendieck theorem for L1 to Hilbert operators would make this map an absolutely summing operator. Its composition with the parametrization would make the identity on the infinite-dimensional l2 sequence space absolutely summing, contradicting its coordinate-vector test.

A bilinear form on a normed vector space is continuous exactly when it is bounded. For the nontrivial direction, continuity at $(0,0)$ gives $r,s>0$ such that $|B(x,y)|\le 1$ whenever $\Vert x\Vert <r$ and $\Vert y\Vert <s$; scaling nonzero $u,v$ into these balls gives $|B(u,v)|\le 4(rs{)}^{-1}\Vert u\Vert \Vert v\Vert$.