I am working through linear algebra problems in Apostol's Calculus, and he has numerous problems that seem to imply that Cauchy-Schwarz holds no matter how the inner product is defined. Then, he has problems where the triangle inequality holds despite alternative definitions of vector norm. This got me thinking, since the proof of the triangle inequality in Apostol relies on Cauchy-Schwarz, that the triangle inequality would hold regardless of how the vector norm is defined (if it involves the dot product).

I then found this response to a question, which states what I was thinking.

Are there any proofs that the Cauchy-Schwarz inequality holds in any inner product space (I looked for some and found none and could not prove it myself)? I've had a semester of algebra (Artin) and some analysis, if there is a proof at such a level of understanding. Intuitive explanations are good, too.

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    $\begingroup$ @AWertheim: That link seems like a perfectly reasonable thing to post as an answer ;) $\endgroup$
    – Jim
    Aug 8, 2013 at 20:42
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    $\begingroup$ @Jim done, and with the advantage of nicer formatting! Cheers :) $\endgroup$ Aug 8, 2013 at 20:44
  • $\begingroup$ "Why does it hold in any space?" could reasonably be taken to mean "Why is there any space in which it holds?". But I doubt that's what you meant. Just changing "any" to "every" would make it completely unambiguous. $\endgroup$ May 23, 2014 at 19:14
  • $\begingroup$ This post is an exceptional proof with good exposition. I probably wouldn't have come up with it myself any faster than the usual discriminant proof but it feels a lot more natural after the fact. Also, a nice formalism for expressing the intuitive idea of "in the worst case". $\endgroup$ May 23, 2014 at 19:29

7 Answers 7


Wikipedia has the one I've usually seen.

Cauchy-Schwarz Inequality

Nothing especially tricky about it, just a really clever setup.


The following is from Cauchy-Schwarz Masterclass by Michael J. Steele. The book also has a small motivation for this proof, and I highly recommend anyone to read it.

For any $u, v$ we have $$(u-v, u-v) \ge 0.$$

Rearranging, we have

$$(u, u) + (v, v) \ge 2 (u, v)$$

or $$||u||^2 + ||v||^2 \ge 2 (u, v).$$

Now, there is a well-known trick to go from an additive inequality (such as the one above) to a multiplicative one (such as C-S). It is called "Normalization". Apply the above inequality to the vectors $u' = u/||u||$, $v' = v/||v||$. Doing so yields

$$1+1 \ge 2(u/||u||, v/||v||)$$

and rearranging yields

$$||u||\cdot ||v|| \ge (u, v)$$ as desired.

  • $\begingroup$ How can you justify the second inequality whenever $(u,v) \in \Bbb C$? $\endgroup$
    – Riaz
    Mar 27, 2023 at 14:56
  • $\begingroup$ @Riaz I guess the C-S inequality itself can't be justified when $(u, v) \in \mathbb{C}$. The C-S inequality says $||u|| \cdot ||v|| \ge (u, v)$, which doesn't make sense unless $(u, v) \in \mathbb{R}.$ So there is nothing to justify. $\endgroup$
    – Ovi
    Jun 18, 2023 at 22:08

The proof on Wikipedia is valid, and is instructive because it uses ideas which are applicable elsewhere in linear algebra. The proof I've usually seen selects vectors u,v and defines

$$f(t) = \langle u + t v , u + t v \rangle$$

and notices that f is nonnegative for all t, by the positive definiteness of the inner product. Expanding out the inner product by bilinearity and combining like terms by symmetry, you get a quadratic function of t:

$$f(t) = \| v \|^2 t^2 + 2 \langle u,v\rangle t + \| u \|^2.$$

(In the complex case you have to do something slightly different.)

This is nonnegative, so its discriminant is nonpositive, in other words:

$$4 \langle u,v\rangle^2 - 4 \| u \|^2 \| v \|^2 \leq 0$$

from which the inequality follows.

An equivalent way of looking at this (for $v \neq 0$) is to find the minimum of $f$. This occurs at $t=-\langle u , v \rangle/\| v \|^2$. So:

$$\langle u , v \rangle^2/\| v \|^2 - 2 \langle u , v \rangle^2/\| v \|^2 + \| u \|^2 = \| u \|^2 - \langle u , v \rangle^2/\| v \|^2 \geq 0$$

A simple rearrangement yields the Cauchy-Schwarz inequality. This is somewhat like the proof on Wikipedia, because $u-\langle u,v \rangle/\| v \|^2 v$ is the projection of $u$ onto the orthogonal complement of the span of $v$.

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    $\begingroup$ Ian, I think the spacing looks better this way. If you disagree, please let me know, and I will roll it back. $\endgroup$ May 23, 2014 at 19:22

Here is an alternative perspective: Cauchy-Schwarz inequality holds in every inner product space because it holds in $\mathbb C^2$. On p.34 of Lectures on Linear Algebra, Gelfand wrote:

Any 'geometric' assertions pertaining to two or three vectors is true if it is true in elementary geometry of three-space. Indeed, the vectors in question span a subspace of dimension at most three. This subspace is isomorphic to ordinary three space (or a subspace of it), and it therefore suffices to verify the assertion in the latter space. In particular the Schwarz inequality -- a geometric theorem about a pair of vectors -- is true because it is true in elementary geometry.

In our case, Gelfand's remark translates to the following. Let $x$ and $y$ be any two vectors in an inner product space $V$ over a field $\mathbb F=\mathbb R$ or $\mathbb C$. Let $S_V=\operatorname{span}\{x,y\}$ and $m=\dim S_V$. By defining an $\mathbb F$-linear map $T$ that maps an orthonormal basis of $S_V$ to an orthonormal basis of a subspace $S_{\mathbb C^2}$ of $\mathbb C^2$ of dimension $m$, one may verify that $T$ is an isometry, i.e. $\langle u,v\rangle_V=\langle Tu,Tv\rangle_{\mathbb C^2}$ for any $u,v\in S_V$. It follows that $|\langle x,y\rangle_V|\le\|x\|_V\|y\|_V$ if and only if $|\langle Tx,Ty\rangle_{\mathbb C^2}|\le\|Tx\|_{\mathbb C^2}\|Ty\|_{\mathbb C^2}$, but the latter is just Cauchy-Schwarz inequality in $\mathbb C^2$.

So, if you can prove that Cauchy-Schwarz inequality holds in $\mathbb C^2$, it must also hold in every inner product space $V$ over $\mathbb R$ or $\mathbb C$, even when $V$ is infinite-dimensional (but $\mathbb C^2$ is not).

Note the subtle difference between this answer and the others. In each of the other answers, a generic proof is given such that the inequality is proved once and for all for every inner product space. Here, we don't prove the inequality directly. Instead, we show that the inequality in a general inner product space follows from its specific version $\mathbb C^2$. How it is actually proved in $\mathbb C^2$ is not our concern. Of course, in order to prove it in $\mathbb C^2$, we may end up using the proofs given in the other answers. However, the point Gelfand made is that if we have learnt a proof of Cauchy-Schwarz inequality in $\mathbb C^2$ in the past, then we don't need to worry whether the proof is generic or not, because the validity of the inequality is automatically generic.

  • $\begingroup$ I am not sure of the significance of this, is this not strong explanation for why inner product spaces are geometric in nature? Why is not this motivation of inner products on abstract spaces seen anywhere else? $\endgroup$ May 16, 2023 at 6:38
  • $\begingroup$ So why is not this explanation more common? $\endgroup$ May 17, 2023 at 20:40

This is an old question but here is an interesting proof that generalizes it a bit. This is adapted from Folland's Real Analysis. Consider some unit scalar $\alpha$ (that is $|\alpha| = 1$) and some $t \in (0,\infty)$. Then,

$$\begin{align*} \langle x -\alpha ty, x-\alpha ty\rangle &= \langle x, x-\alpha ty\rangle + \langle -\alpha ty, x-\alpha ty\rangle \\ &= \langle x,x\rangle - \left(\overline{t} \langle x, \alpha y\rangle +t \langle \alpha y,x\rangle\right) + \underbrace{\alpha\overline{\alpha}}_{=|\alpha|^2=1} t^2 \langle y,y\rangle \\ &= \langle x,x\rangle - \underbrace{\left(\overline{t \langle x, \alpha y\rangle} + t \langle x,\alpha y\rangle\right)}_{2t\operatorname{Re} \langle x,y\rangle} + t^2 \langle y,y\rangle \\ &= \|y\|^2 t^2 - 2\operatorname{Re} \langle x,\alpha y\rangle t + \|x\|^2. \end{align*}$$

Now, this is a quadratic in $t$, and we observe that it opens upward and is non-negative. Therefore, consider its discriminant, $b^2 -4ac \le 0$.

We have $b = 2\operatorname{Re} \langle x,\alpha y\rangle$. However, since $\alpha = \frac{z}{\|z\|} = \operatorname{sgn} z$ for some $z$, we have $\langle x, \alpha y\rangle = \langle \alpha y, x\rangle = |\langle x, y\rangle|$; hence $\operatorname{Re}\langle x,\alpha y\rangle = |\langle x, y\rangle|$.

Therefore, we obtain

$$4|\langle x,y\rangle|^2 \le 4\|x\|^2\|y\|^2 \implies |\langle x,y\rangle| \le \|x\|\|y\|.$$

  • $\begingroup$ I enjoy this because there is something fun about a proof for a concept that generalizes to fairly advanced structures using techniques developed in middle-school. $\endgroup$
    – Emily
    May 3, 2015 at 15:45
  • $\begingroup$ how does $\langle x, \alpha y\rangle=\langle\alpha y, x\rangle=|\langle x, y\rangle|$ hold? $\endgroup$
    – user149418
    Feb 15, 2018 at 16:28

Let $t=\operatorname{sign}(\langle v,w\rangle)$, then $t\in \{-1,0,1\}$ and by triangle inequality, $|v+tw|\leq |v|+|t||w|\leq |v|+|w|$, thus $$|v + tw|^2=|v|^2+2t\langle v,w\rangle+|w|^2\leq |v|^2+2|v||w|+|w|^2.$$ Hence $$t\langle v,w\rangle=|\langle v,w\rangle|\leq |v||w|.$$

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    $\begingroup$ Try proving the triangle inequality without Cauchy-Schwarz. $\endgroup$
    – Unit
    Dec 29, 2014 at 18:32

Indeed it holds in any sesquilinear form in any vector space not even (necessarily) normed space. The proof is trivial.


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