Let \( V \) be an inner product space and \( u, v \in V \). Then
\[ |\langle u, v \rangle| \leq \|u\| \, \|v\|, \]with equality holding if and only if \( u \) and \( v \) are linearly dependent.
Proof
Assume \(v\neq 0\). Otherwise, the inequality holds trivially. Let
\[ z=u-\frac{\langle u, v\rangle}{\langle v, v\rangle}v. \]The vector \(z\) is orthogonal to the vector \(v\), and thus, by the Pythagorean theorem and by the identity
\[ u=\frac{\langle u, v\rangle}{\langle v, v\rangle}v + z, \]we deduce
\[ \lVert u\rVert^2 = \frac{\lvert \langle u, v\rangle^2\rvert}{\lVert v\rVert^2} + \lVert z\rVert^2 \ge \frac{\lvert \langle u, v\rangle^2\rvert}{\lVert v\rVert^2}. \]The Cauchy-Schwartz inequality follows by multiplying with \(\lVert v\rVert^2\) and taking the square root.
The above inequality is an equality if and only if \(z=0\). This is the case if \(u\) and \(v\) are linearly independent.
Remarks
- The Cauchy-Schwartz inequality implies continuity of the inner product .