I am wondering whether the Cauchy-Schwarz inequality does hold when absolute value is not considered for the LHS.

Let me explain: In standard Cauchy-Schwarz we have:

$| \langle \textbf{u},\textbf{v}\rangle |\;\leq \|\textbf{u}\| \cdot \|\textbf{v}\| $

But will this hold if no absolute value is taken for the left hand side?

My thoughts:

Since $| \langle\textbf{u},\textbf{v}\rangle |\; = \|\textbf{u}\| \cdot \|\textbf{v}\| \cos(\alpha) $

we have that since $\cos(\alpha) \leq 1$, then the inequality:

$ \langle\textbf{u},\textbf{v}\rangle \;\leq \|\textbf{u}\| \cdot \|\textbf{v}\| $, should hold.

May I get your opinion, thoughts? please!

  • $\begingroup$ Are you allowing complex values for $\left<u,v\right>$? $\endgroup$ Mar 1, 2019 at 6:55
  • $\begingroup$ Thanks for your reply. Not really, I assume $u$ and $v$ are real. (by the way, what is the $\LaTeX$ sintax for inner product? $\endgroup$
    – kentropy
    Mar 1, 2019 at 6:56

1 Answer 1


Note that $\langle u,v\rangle\leqslant\bigl\lvert\langle u,v\rangle\bigr\rvert\leqslant\lVert u\rVert.\lVert v\rVert$.

  • $\begingroup$ Thanks. Yeah I agree, it makes sense. $\endgroup$
    – kentropy
    Mar 1, 2019 at 7:24

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