My professor tried to prove the Cauchy-Schwarz inequality in a weird way which I can't understand and I couldn't find an example where it's replicated. I say "tried" because he didn't finish it and gave it out to finish as an exercise.

So for $\forall x,y$ in an inner product space, and $\forall t \in \mathbb{R}$:

$$0\leq\langle x+ty,x+ty\rangle = \langle x,x\rangle+t \langle x,y\rangle+t \langle y,x\rangle+t^2 \langle x,x\rangle = \langle x,x\rangle + 2t\Re{(\langle x,y\rangle})+ t^2\langle y,y\rangle$$

Notice that $t \in \mathbb{R}$ so I left out conjugation, on purpose. The rest of the equations are coming from using the inner product axioms.

There are two cases cases, the first is that: $$y = 0 \implies \langle x,y \rangle = 0$$ in which case, the inequality is trivial.

The seoncd is that $y \neq 0$. At this point, the polynomial $\langle x,x\rangle + 2t\Re{\langle x,y\rangle}+t^2\langle y,y \rangle$ is a quadratic polynomial which is greater than or equal to zero $\forall t \in \mathbb{R}$. By this we can conclude that the discriminant of the polynomial is less than or equal to zero, which means: $$4\Re{(\langle x,y\rangle)}^2-4\langle x,x\rangle^2\langle y,y\rangle^2 \leq 0 \implies 4\Re{\langle x,y\rangle}^2 \leq 4\langle x,x\rangle^2\langle y,y \rangle^2$$

At this point he said the rest is left as an exercise. I tried to finish it using trivial means, writing the real part in trigonometric form, and such, but no success.

What I don't understand is why did we choose $t \in \mathbb{R}$ when we could choose $t\in \mathbb{C}$ and then choose it's value to help us get the desired result. I have no clue how to progress from here and the more I think about the more I believe that it's probably very easy to solve, I just simply don't see it.


For the complex $\left<x,y\right>$, find some complex $z$ with $|z|=1$ such that $z\left<x,y\right>=|\left<x,y\right>|$, then $\text{Re}z\left<x,y\right>$ is real and hence \begin{align*} |\left<x,y\right>|^{2}&=(\text{Re}z\left<x,y\right>)^{2}\\ &=(\text{Re}\left<zx,y\right>)^{2}\\ &\leq\left<zx,zx\right>\left<y,y\right>\\ &=|z|^{2}\left<x,x\right>\left<y,y\right>\\ &=\left<x,x\right>\left<y,y\right> \end{align*}

  • $\begingroup$ Oh my... That's so easy it hurts. Thanks, very helpful, you just saved my night. $\endgroup$ – Levente Kovács Apr 17 '18 at 0:43
  • $\begingroup$ Okay, this is awkward but revisiting this proof something is not quite clear. Why is it possible to solve the $z\left<x,y\right>=\text{Re}\left<x,y\right>$ equation for every x and y? What if the dot product in $\mathbb{C}$ is simply, for example $i$? Then we would get that $|\left<x,y\right>| = |\text{Re}\left<x,y\right>|$, but $|\left<x,y\right>| = 1$, since it equals to $i$, but $|\text{Re}\left<x,y\right>| = 0$. Am i missing something? $\endgroup$ – Levente Kovács May 10 '18 at 18:36
  • 1
    $\begingroup$ I have made some mistake, now it goes through. $\endgroup$ – user284331 May 11 '18 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.