Proof of the Cauchy Shwarz inequality

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My mathematical methods for theoretical physics course recently began looking at linear vector spaces. We defined the Banach and Hilbert Spaces and proved the Cauchy-Shwarz Inequality. There's one step in this proof that I cant really follow (in red):

consider: w=x+uy (i'll drop the emboldening of vectors now to save time)

0 <= <x,x> + <uy,x> + <x,uy> + <uy,uy>

0 <= <x,x> + u*<y,x> + u<x,y> + u*u<y,y>

Choose u = <y,x>/<y,y>

0 <= <x,x> + (<y,x>/<y,y>)*<y,x> + (<y,x>/<y,y>)<x,y> + (<y,x>/<y,y>)2<y,y>

some manipulation here concludes the proof.

My question is that by choosing a specific u are we not losing generality?
 

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  • #2
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On the contrary, the inequailty is true for all u, including the chosen one.
Moreover, it should be w=x-uy instead.
 
  • #3
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The way I remember seeing it is that you want to choose u to minimize the RHS of

0 <= <x,x> + u*<y,x> + u<x,y> + u*u<y,y>

Then any other u will make the RHS larger and thus still satisfy the inequality.
So take the derivative wrt u* to get

0 <= 0 + <y,x> + 0 + u<y,y>

==> u = - <y,x> / <y,y>

(You can check that extremizing wrt u gives a consistent result)

==> 0 <= <x,x> - 2<y,x><x,y>/<y,y> + <y,x><x,y>/<y,y>

==> <y,x><x,y> <= <x,x><y,y>
 
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  • #4
Fredrik
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0 <= <x,x> + <uy,x> + <x,uy> + <uy,uy>

0 <= <x,x> + u*<y,x> + u<x,y> + u*u<y,y>

Choose u = <y,x>/<y,y>
You need to choose u to minimize the right-hand side. To do this, you need to choose Arg u=-Arg (u<x,y>)+π, because this makes u<x,y> real and negative. (Note that u*<y,x> + u<x,y>=2 Re (u<x,y>).) Then you find the value of |u| that makes the right-hand side as small as possible. It looks like you're trying to do both of these things at once. There's nothing wrong with that, but it makes it more difficult to see that we're just trying to minimize the right-hand side. You also got the sign wrong.

My question is that by choosing a specific u are we not losing generality?
No. The proof goes like this:
Let x and y be arbitrary vectors. Then for any complex number u, we have 0≤ <x+uy,+uy>. In particular, this inequality holds when u=(the value that minimizes the right-hand side), and in that case, it can be simplified to the inequality we're trying to prove.​
We haven't lost any generality since x and y are still arbitrary.
 
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  • #5
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Ah, I get it. Yeah, sorry for the missing sign. Thanks a lot for your help, guys!
 
  • #6
mathwonk
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one clever proof introduces a variable t, and looks at (a+tb).(a+tb) = |a+tb|^2 and uses that this is always non negative. I.e. the graph is a parabola that does not cross the x axis except to touch it at most once. In particular it has one or no real roots. Since it is a quadratic equation in t, that means the discriminant of the equation is non positive. that is the cauchy schwartz inequality, as i recall, unless i screwed it up. that looks like the same proof above, which minimizes a quadratic.
 
  • #7
Fredrik
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I actually saw that proof (the one mathwonk is talking about) for the first time less than a week ago. (Note that my inner product is antilinear in the first variable and linear in the second).

[tex]
\begin{align*}\label{eq:cbst}
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 y,y\rangle\nn\\
&= \|x\|^2+2\operatorname{Re}(t\langle x,y\rangle)+|t|^2\|y\|^2
\end{align*}
[/tex]

The choice [itex]\operatorname{Arg} t=-\operatorname{Arg}(t\langle x,y\rangle)+\pi[/itex] turns this into

[tex]0\leq\|x\|^2-2|t||\langle x,y\rangle|+|t|^2\|y\|^2[/tex]

Now consider the polynomial p, defined by

[tex]p(z)=\|y\|^2 z^2-2|\langle x,y\rangle|z+\|x\|^2[/tex]

Its roots are

[tex]\frac{|\langle x,y\rangle|}{\|y\|^2}\pm\frac{1}{\|y\|^2}\sqrt{\frac{|\langle x,y\rangle|^2}{\|y\|^2}-\|x\|^2}[/tex]

The clever trick is to note that since p is a second degree polynomial and p(z)≥0 when z is real, p can't have more than one real root, but if the quantity under the square root above is positive, it has two! So, we must have

[tex]\frac{|\langle x,y\rangle|^2}{\|y\|^2}-\|x\|^2\leq 0[/tex]
 

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