= Bounded bilinear form
{title2=$|B(u,v)|\leq C\|u\|\|v\|$}
A <bilinear form> $B:X\times X\to\mathbb R$ on a real <normed vector space> is bounded when one constant $0\leq C<\infty$ satisfies
$$
|B(u,v)|\leq C\|u\|\|v\|\qquad(u,v\in X).
$$
This 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>.<Introduction to Finite Element Methods by Christian Clason>[§2.2, Theorem 2.7(ii): continuity bound]{ref}
For example, on $\mathbb R^2$ with the <Euclidean norm>, the <bilinear forms>
$$
B_-(u,v)=u_1v_1-u_2v_2,\qquad B_0(u,v)=u_1v_1
$$
both 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 $B(u,v)=u^TAv$ using its <matrix of a bilinear form>. The <Cauchy-Schwarz inequality> gives $|u^TAv|\leq\|A\|_2\|u\|_2\|v\|_2$, with $\|A\|_2$ 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 set
$$
B(u,v)=\sum_{n\geq1}n u_n v_n.
$$
Each pairing is a finite sum, so this is a well-defined <bilinear form>. For the <basis vector> $e_n$, however, $\|e_n\|_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a <coercive bilinear form> with $\alpha=1$, since $\sum n v_n^2\geq\sum v_n^2$. 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> $T:X\to X'$ by $(Tu)(v)=B(u,v)$, where $X'$ is the <continuous dual space>. Its <operator norm> is the smallest admissible $C$:
$$
\|T\|=\sup_{u\ne0,\,v\ne0}\frac{|B(u,v)|}{\|u\|\|v\|}.
$$
This converts an upper bound for pairings into an upper bound for a <linear map> taking an unknown <vector> to a <linear functional>.
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