Some musings on Cauchy-Schwarz, Part Vb

-

 Recall our claim from Part I

Cauchy-Schwarz in complex inner product spaces can be deduced from Cauchy-Schwarz in real inner product spaces. This is essentially because every complex inner product space can also be regarded as a real inner product space with the same norm, but where the real inner product is the real part of the complex inner product. The last bit of the deduction uses a fairly standard "rotation trick": for any complex number \(z\), we have 

\[|z|=\max\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\}\,.\qquad(*)\]

Let's check $(*)$ first.

Let $z\in\C$, and set

\[A=\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\} \subseteq \R\,.\]

 Clearly $A$ is non-empty.

 Let $\alpha \in \C$ with $|\alpha|=1$. Then certainly we have

\[\Re(\alpha z) \leq |\alpha z| = |z|\,.\]

Thus we have

\[\sup A \leq |z|\,.\]

We now need to prove that equality holds, and that the supremum is actually a maximum. For this we just need to show that $|z| \in A$.

This is trivial if $z=0$, because then we have $A=\{0\}$.

Otherwise, consider $\alpha=|z|/z$. Then $|\alpha|=1$,  and $\alpha z = |z|$, so $\Re(\alpha z)=|z|$ too. This shows that $|z| \in A$, as required.

Equation (*) is a very standard "rotation trick", used repeatedly in analysis.

Here is another way to think of it. Multiplying by a complex number $\alpha$ with $|\alpha|=1$ just rotates $z$ around the origin, and (for fixed $z \neq 0$) as $\alpha$ varies,  the path traced out by $\alpha z$ is a circle centred on $0$ with radius $|z|$. You may prefer to think of $\alpha$ as $\e^{\i\theta}$, where $\theta\in \R$ is the angle we rotate by. When we take the real part we are applying the (orthogonal) projection from $\C$ onto the real axis, and $A$ is the image of our circle under this projection, and this image is clearly just the closed interval $\left[\,{-|z|},|z|\,\right]$. Equation $(*)$ now says that $\max(\left[\,{-|z|},|z|\,\right])=|z|$, which is completely obvious.

Returning to complex inner product spaces, we will work with the convention that complex inner products are linear in the second variable. (But you can easily modify the argument to deal with the alternative convention.)

Let $V$ be a complex inner product space, with (complex) inner product $\langle\cdot,\cdot\rangle_C$ and associated norm $\|\cdot\|$. Then, if you keep the same addition but only consider multiplication by real scalars, $V$ is also a real inner product space with the same norm, but using the real inner product $\langle\cdot,\cdot\rangle_R$ given by

\[\langle\vx,\vy\rangle_R = \Re(\langle\vx,\vy\rangle_C)\,.\]

We know that Cauchy-Schwarz holds for $V$ when regarded as a real inner product space, which tells us that, for all $\vx,\vy \in V$, we have

\[|\Re(\langle\vx,\vy\rangle_C)| = |\langle\vx,\vy\rangle_R| \leq \|\vx\| \|\vy\|\,,\]

with equality if and only if $\vx$ and $\vy$ are linearly dependent over $\R$.

For the complex version, we don't want the real part on the left hand side, and we want to show that equality then holds if and only if $\vx$ and $\vy$ are linearly dependent over $\C$ instead.

For this, we use the rotation trick. By $(*)$, we can choose $\alpha\in\C$ with $|\alpha|=1$ such that 

\[\alpha \langle\vx,\vy\rangle_C = |\langle\vx,\vy\rangle_C|\,.\]

Then (because our complex inner product is linear in the second variable),

\[\langle\vx,\alpha \vy\rangle_C = |\langle\vx,\vy\rangle_C|\,,\]

and so

\[|\langle\vx,\alpha\vy\rangle_R| =|\Re(\langle\vx,\alpha \vy\rangle_C)|=|\langle\vx,\vy\rangle_C|\,.\]

By the real version of Cauchy-Schwarz, we have

\[|\langle\vx,\alpha\vy\rangle_R| \leq \|\vx\| \|\alpha \vy\| = \|\vx\| \|\vy\|\,,\]

and so

\[|\langle\vx,\vy\rangle_C| \leq \|\vx\| \|\vy\|\,,\]

as required.

When does equality hold here? Well, with $\alpha$ as above, if equality holds then $\vx$ and $\alpha \vy$ must be linearly dependent over $\R$, and hence $\vx$ and $\vy$ are linearly dependent over $\C$. Conversely, if $\vx$ and $\vy$ are linearly dependent over $\C$ (i.e., one is a complex scalar multiple of the other), you can easily confirm that equality holds. So equality holds if and only if $\vx$ and $\vy$ are linearly dependent over $\C$. 

This is not the end of the story, as mentioned in Part IVb. But I will pause these posts for now, as I have to meet some deadlines!

Comments

Post a Comment

Popular posts from this blog

A rule of thumb for when to use the Ratio Test

-
The Ratio Test is a powerful test for both sequences and series of non-zero real numbers. It comes in various forms, but here is one commonly used version. Theorem (Ratio Test) Let $(x_n)_{n\in\N}$ be a sequence of non-zero real numbers. Suppose that \[\frac{|x_{n+1}|}{|x_n|}\to L \text{ as } n\to\infty\,,\] where $L$ is either a non-negative real number or $+\infty$. (a) If $L>1$, then $|x_n| \to +\infty$ as $n \to + \infty$, the sequence $(x_n)$ is divergent, and the series $\displaystyle \sum_{n=1}^\infty x_n$ diverges. (b) If $L \in [0,1)$, then the series $\displaystyle \sum_{n=1}^\infty x_n$ is absolutely convergent (and hence convergent), and the sequence $(x_n)$ converges to $0$. (c) If $L=1$, then the Ratio Test is inconclusive , and you need to use a different test . (The Ratio Test tells you nothing in this case about the convergence or otherwise of either the sequence or the series.) Notes Here (b) does not tell you anything about the value of the sum of the series ...
Read more

The Extreme Value Theorem: a bootstrap approach

-
At the very end of last week I was teaching the first-year mathematics students about the Extreme Value Theorem. Here is a reminder of what the theorem says. Theorem (Extreme Value Theorem) Let $a,b$ be real numbers with $a<b$, and let $f:[a,b]\to\R$ be continuous. Then there exist points $c$ and $d$ in $[a,b]$ such that, for all $x \in [a,b]$, we have $f(c)\leq f(x) \leq f(d)$. Here we have \[f(c)=\min\{f(x):x\in[a,b]\}=\inf\{f(x):x\in[a,b]\}\] and  \[f(d)=\max\{f(x):x\in[a,b]\}=\sup\{f(x):x\in[a,b]\}\,.\] Unfortunately I only had a few minutes left for this, ran out of time, and didn't finish the proof. As I am on a very tight schedule this year, I had to refer the students to the the typed notes for the rest of the proof. On the other hand, I did finish the first part of the proof, showing that a continuous real-valued function $f$ on a closed and bounded interval $[a,b]$ is always bounded above, i.e., the set $f([a,b])$ is bounded above. In this post I want to see how you ca...
Read more

An introduction to the Hahn-Banach extension theorem: Part VI

-
In this post we sketch a second proof of the Hahn-Banach Extension Theorem for continuous linear functionals on real normed spaces, this time using transfinite induction. We omit some of the details, which the reader can fill in. In particular, some of the details are similar to those involved in proving the claim made in Part V (when we obtained an upper bound for a chain). For a non-zero ordinal $\gamma$, recall that, with ordinal interval notation, we have $\gamma=[0,\gamma)$. That is, $\gamma$ is equal to the set of all ordinals which are strictly less than $\gamma$. However, we use ordinal interval notation throughout. If you prefer to use $\gamma$ instead of $[0,\gamma)$ that is fine, of course! Looking at the Wikipedia page on the Hahn-Banach Theorem , apparently both Hahn and Banach (independently) used transfinite induction in their proofs, rather than Zorn's Lemma. Theorem (Hahn-Banach Extension Theorem for linear functionals on normed spaces over $\R$)  Let $(E,\jnorm)$ ...
Read more