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We know the Cauchy–Schwarz inequality for inner product space such that

\begin{align} |\langle u,v \rangle|^2 \le \langle u,u \rangle \langle v,v \rangle \end{align} Then if we induce the norm by $\|u\| := \sqrt{\langle u,u \rangle}$ we obtain \begin{align} |\langle u,v \rangle| \le \| u \| \| v \| \end{align}

On the other hand, for a continuous bilinear form $b(u,v)$ that is bounded we have \begin{align} b(u,v) \le C \| u \| \| v \| \end{align} which is very similar to the Cauchy–Schwarz inequality except a constant $C$.

I already know the inner product can been considered as a symmetric nonnegative bilinear form. While I do not yet find but I believe there are some inherent connections between the Cauchy–Schwarz inequality and the continuity/boundedness inequality. Can someone give me an answer?

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  • $\begingroup$ Note that Cauchy-Schwarz is used to prove that the induced norm satisfies triangle inequality. On the other hand, for inner product space $V$ we have that $V$ embeds into dual space $V^*$ via $v\mapsto \langle -, v\rangle$. Notice that Cauchy-Schwartz makes sure that these functionals are bounded. If moreover $V$ is Hilbert, Riesz representation theorem tells us that this embedding is isometric anti-isomorphism. $\endgroup$ – Ennar May 10 '18 at 13:30

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