I am looking for a proof of the following theorem: Given a vector $\textbf{a}$ in an inner product space, the mapping $\sqrt{\langle\textbf{a},\textbf{a}\rangle}$ is a norm of $\textbf{a}$.

This is shown by considering the axioms of the inner product ${\langle\textbf{a},\textbf{a}\rangle}$ and showing that $\sqrt{\langle\textbf{a},\textbf{a}\rangle}$ satisfies the axioms of a norm. The positive definite axiom holds for both norms and inner products. The absolute homogeneity axiom for a norm can also be shown in a few steps.

My problem is with proving that the triangle inequality holds for a function $f(\textbf{a}+\textbf{b})=\sqrt{\langle\textbf{a}+\textbf{b},\textbf{a}+\textbf{b}\rangle}$ such that $f(\textbf{a}+\textbf{b})\leq f(\textbf{a})+f(\textbf{b})$. The proofs I've seen use the Cauchy-Schwarz inequality to prove this step. My problem is that all proofs of Cauchy-Schwarz itself contain a step that assumes that $\sqrt{\langle\textbf{a},\textbf{a}\rangle} = ||\textbf{a}||$. But this is the relation we are trying to prove in the first place.

So is there a general proof of the Cauchy-Schwarz inequality that doesn't assume $\sqrt{\langle\textbf{a},\textbf{a}\rangle} = ||\textbf{a}||$? Or if such a proof doesn't exist: is there a proof of the norm-inner product equivalence theorem that doesn't seem to assume itself in its own proof? Unless I am misunderstanding something.

  • 1
    $\begingroup$ Check those proofs again. What they usually use is that $\langle\textbf{x},\textbf{x}\rangle \ge 0$ for all vectors $\textbf{x}$. $\endgroup$
    – Martin R
    Jun 11 at 7:55
  • 3
    $\begingroup$ "all proofs of Cauchy-Schwarz [assume] $\sqrt{\langle\textbf{a},\textbf{a}\rangle} = ||\textbf{a}||$. But this is the relation we are trying to prove in the first place." -- You must be misunderstanding something crucial, because this is not the relation that the Cauchy-Schwarz inequality cares about or tries to prove. $\endgroup$ Jun 11 at 7:58
  • 3
    $\begingroup$ First and foremost, $\sqrt{\langle a,a \rangle} = \|a\|$ is, in the broader general context for all inner product spaces, the definition of the norm (the norm we say is "induced" by the inner product). That it coincides with a certain formula in $\mathbb{R}^n$ is certainly nice, of course, but the Cauchy-Schwarz inequality holds outside of that context too. $\endgroup$ Jun 11 at 7:58
  • $\begingroup$ @Martin R Could you link such a proof? The only one I could find that doesn't assume norm-inner product equivalence resorts to using Pythagoras, which I don't believe is general to all vector spaces. $\endgroup$
    – pll04
    Jun 11 at 9:00
  • $\begingroup$ @pll04 You have ignored the other comments. $\langle a,a\rangle=\|a\|^2$ is true: usually by definition, and in the Euclidean case, if you like, it is true by a very easy computation of the scalar product. This is not an "assumption", it is a (trivial/definitional) fact $\endgroup$
    – FShrike
    Jun 11 at 9:06

1 Answer 1


There's a bit of a logical mixup here.

You want to show, $x\mapsto\sqrt{\langle x,x\rangle}$ is a norm on any inner product space). The only axiom of a norm that is not immediate is the triangle inequality.

So you want to bound: $$\sqrt{\langle x+y,x+y\rangle}\le\sqrt{\langle x,x\rangle+\langle y,y\rangle+2\langle x,y\rangle}\le\sqrt{\langle x,x\rangle+\langle y,y\rangle+2\sqrt{\langle x,x\rangle\cdot\langle y,y\rangle}}$$Using the CBS inequality, and then from there you just observe: $$\sqrt{\langle x,x\rangle+\langle y,y\rangle+2\sqrt{\langle x,x\rangle\cdot\langle y,y\rangle}}=\sqrt{\langle x,x\rangle}+\sqrt{\langle y,y\rangle}$$Concluding the triangle inequality.

Now you object, because the CBS inequality is often stated as: $$\langle a,b\rangle^2\le\|a\|^2\|b\|^2$$Where $\|\cdot\|$ is "already a norm", or something. But really, when stating the CBS inequality and when proving it, the symbol "$\|\cdot\|$" is simply a placeholder for the symbol $\sqrt{\langle,\rangle}$ and the fact that $\|\cdot\|$ - so defined - is a genuine norm is not used.

So there is no problem at all.

The standard proof of the CBS inequality (real case): (this aspect of the post is very much a duplicate, save for the avoidance of norm notation - see here and here)

Fix $a,b$ and consider the map: $$f:\Bbb R\ni t\mapsto\langle a+tb,a+tb\rangle\in\Bbb R$$

We have: $$f(t)=t^2\langle b,b\rangle+2t\langle a,b\rangle+\langle a,a\rangle$$But also: $$f\ge0$$

So the discriminant must be nonpositive, i.e.: $$(2\langle a,b\rangle)^2-4\langle a,a\rangle\cdot\langle b,b\rangle\le0$$And then: $$\langle a,b\rangle^2\le\langle a,a\rangle\cdot\langle b,b\rangle$$Follows. Then: $$|\langle a,b\rangle|\le\sqrt{\langle a,a\rangle\cdot\langle b,b\rangle}$$Follows. This inequality can be used in the above proof that $x\mapsto\sqrt{\langle x,x\rangle}$ is a norm. There is absolutely no problem, and nowhere was the $\|\cdot\|$ notation needed or used.

If the space is a complex inner product space we use a very similar approach. Fix $a,b$ as before and define $f$ in the same way. Let $g:\Bbb C\to\Bbb R$ be the map: $$\lambda\mapsto|\lambda|^2\cdot\langle a,a\rangle+2|\lambda|\cdot|\langle a,b\rangle|+\langle b,b\rangle$$

Because: $$\begin{align}g(\lambda)&\ge|\lambda|^2\cdot\langle a,a\rangle+2\cdot\Re(\lambda\cdot\langle a,b\rangle)+\langle b,b\rangle\\&=\langle \lambda a+b,\lambda a+b\rangle^2\\&=f(|\lambda|)\\&\ge0\end{align}$$

For all $\lambda\in\Bbb C$, we conclude that the polynomial: $$\Bbb R\ni x\mapsto\langle a,a\rangle\cdot x^2+2|\langle a,b\rangle|\cdot x+\langle b,b\rangle\in\Bbb R$$Is nonnegative and so has nonpositive discriminant.

This discriminant is basically the same one as before, except with a modulus sign - you get: $$4|\langle a,b\rangle|^2-4\langle a,a\rangle\cdot\langle b,b\rangle\le0\implies|\langle a,b\rangle|^2\le\langle a,a\rangle\cdot\langle b,b\rangle$$

  • $\begingroup$ Thank you! But how can we extend your CBS proof to the complex case? Then, in the second to last expression we would get Re(a,b)^2 =< (a,a)(b,b) and we know that |(a,b)|^2 = Re(a,b)^2 + Im(a,b)^2. Therefore Re(a,b)^2 =< |(a,b)|^2 and Re(a,b)^2 =< (a,a)(b,b). However, this does not necessarily imply that |(a,b)|^2 =< (a,a)(b,b)? $\endgroup$
    – pll04
    Jun 11 at 11:37
  • $\begingroup$ @pll04 You're welcome. Updated to account for the complex case $\endgroup$
    – FShrike
    Jun 11 at 12:45
  • $\begingroup$ @pll04 from the inequality $|\text{Re}(\langle a,b\rangle)|\leq \|a\|\|b\|$, you can “amplify” to get the usual Cauchy-Schwarz, because this inequality has a “phase imbalance”. Simply replace $a$ by $e^{i\theta}a$, then choose the angle to offset the argument of the original inner product. See here for more details (where you also see that the full strength of the inner product is not needed (positive-definiteness or even non-degeneracy is not needed)). $\endgroup$
    – peek-a-boo
    Jun 11 at 16:16
  • $\begingroup$ @peek-a-boo Can you explain the last part of the proof you linked? How do we justify the inequality Re(e^i(theta) * phi(x,y)) >= |phi(x,y)| (which we need to make the last step)? $\endgroup$
    – pll04
    Jun 12 at 10:43
  • $\begingroup$ @pll04 I already said to choose $\theta$ to offset the argument. In more detail: $\phi(x,y)$ is a complex number, so we can write it as $\phi(x,y)=|\phi(x,y)|e^{i\theta_0}$ for some $\theta_0\in\Bbb{R}$ (if $\phi(x,y)=0$ then every $\theta_0\in\Bbb{R}$ works. If $\phi(x,y)\neq 0$, then $\theta_0$ is unique up to addition of integer multiples of $2\pi$). Now, choose $\theta=-\theta_0$. Then, $e^{i\theta}\phi(x,y)=|\phi(x,y)|$, so the real part of this is simply $|\phi(x,y)|$. This is called “amplifying the phase imbalance” (read the Terry Tao page for more). $\endgroup$
    – peek-a-boo
    Jun 12 at 17:06

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