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!