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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.

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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*}

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  • $\begingroup$ 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
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    $\begingroup$ I have made some mistake, now it goes through. $\endgroup$ – user284331 May 11 '18 at 23:25

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