# Show the Cauchy-Schwarz inequality holds on a Hilbert space

How would one go about showing this? Its a question in one of the workbooks but it doesn't provide an answer. Any help would be appreciated.

• Note that for all $t\in\mathbb{R}$, you have $\lVert x-ty\rVert^2 \geqslant 0$. Choose $t$ so that the inequality drops out. – Daniel Fischer Mar 10 '14 at 12:20
• Gowers has some interesting remarks about the inequality and its proofs on his website. – Did Mar 10 '14 at 12:33
• Possible duplicate of math.stackexchange.com/questions/436559/… – littleO Mar 10 '14 at 19:09

Somehow, on the whole internet, it seems that the simplest proof of Cauchy- Schwarz has yet to be recorded. At least I couldn't find it after several minutes of searching... The most prominent is certainly the proof mentioned by Daniel Fischer in this comment above, but that always seemed quite contrived to me. Here is the best'' proof imho:

let $x,y$ be unit vectors.

Then $\langle x-y,x-y \rangle = |x|^2-2\langle x,y\rangle+|y|^2 \geq 0$

so $\langle x,y \rangle \leq 1$

Now for any two nonzero vectors, $x,y$ (if one is $0$ the result is trivial), we have that

$\left\langle \frac{x}{|x|},\frac{y}{|y|}\right\rangle \leq 1$ by the result above.

So $\langle x,y \rangle \leq |x||y|$

Of course, we also need to show that $\langle x,y \rangle \geq -|x||y|$, but I will leave it to you to see how to modify the argument to obtain this inequality.

• You have to show that $\langle x,y\rangle\ge-1$ in case both vectors are unit vectors, too. – Michael Hoppe Mar 10 '14 at 14:44
• @MichaelHoppe I hope that it is clear how to establish that in much the same way...Didn't want to give it all away to OP. – Steven Gubkin Mar 10 '14 at 16:03
• Well done, but in my opinion a hint about the second part would not be unnecessary. – Michael Hoppe Mar 10 '14 at 17:26
• I will add it to the post. – Steven Gubkin Mar 10 '14 at 18:34
• It's worth noting that this proof does not work as nicely for complex Hilbert spaces, in which $||x||^2 + ||y||^2 - 2\mathfrak{Re}\langle x,y \rangle \geq 0$. – Ben Bray Mar 11 '18 at 3:10

I am not sure what is the "best" proof of this famous inequality, but I found the following remark helpful, at least conceptually.

For a 2 dimensional Hilbert space, i.e. the usual Euclidean plane of highschool math, the inequality is quite elementary and intuitive, by some drawing, or even working in coordinates, it is straighfword to show that $(ac+bd)^2\leq (a^2+b^2)(c^2+d^2)$.

Now the remark is that for a general pair of vectors in a whatever dimension Hilbert space, you can consider the two dimensional subspace spanned by these two vectors (supposing they are not linearly dependent, in which case there is nothing to prove). We are reduced to the previous case.

In other words, it is really a plane geometry inequality, nothing more.

As the Gramian matrix of $(x,y)$, namely $G(x,y):=\begin{pmatrix}\langle x,x\rangle & \langle x,y\rangle\\ \langle y,x\rangle & \langle y,y\rangle\end{pmatrix}$, is well known to be positiv-semidefinit, we know that $\det\bigl(G(x,y)\bigr)\ge0$ and equality holds iff $x\parallel y$.

• Cauchy-Schwarz is probably even more "well known" than this. – Did Mar 10 '14 at 17:27
• Very nice, but uses a lot of machinery. It is elementary to show that it is positive semidefinite, but then I think we need the real spectral theorem to make the claim that the determinant is positive. And we must develop the theory of determinants, eigenvectors, bilinear forms, etc all before this. Still very neat! – Steven Gubkin Mar 10 '14 at 18:46
• Thanks. A student discoverin Hilbert spaces is supposedto know of that “machinery”. And why not use it as in the following proof for $\sqrt[2]{3}$ is irrational. Suppose $2=p^3/q^3$. Then $q^3+q^3=p^3$ which is impossible due to a result of Wiles. – Michael Hoppe Mar 11 '14 at 8:11