Why $\langle y,x\rangle+\langle x,y\rangle=2\mathrm{Re}\langle x,y\rangle$? And the rules of using absolute value, inner production and norm?

Question

Let V be an inner product space over F, x,y∈V. In the proof of triangle inequality, my textbook uses $$\|x+y\|^2 = \langle x,x \rangle + \langle y,x \rangle + \langle x,y \rangle + \langle y,y \rangle = \|x\|^2 + 2\mathrm{Re}\langle x,y \rangle + \|y\|^2$$ $$\leq \|x\|^2 + 2|\langle x, y \rangle| + \|y\|^2$$ directly. Here $\langle \cdot, \cdot \rangle$ may not be the standard inner product.

I tried to use $x=a+bi$ and $y=a'+b'i$ to verify if $\langle y,x \rangle + \langle x,y \rangle = 2\mathrm{Re} \langle x,y \rangle$, and it turned out to be $2\langle a,a'\rangle+2\langle b,b'\rangle = 2\mathrm{Re}\langle x, y \rangle$.

Then I wonder what is the real part of $\langle x, y \rangle$ exactly?

For the next step, how come $\mathrm{Re}\langle x,y \rangle \leq |\langle x, y \rangle|$? What is $|\langle x, y \rangle|$ then?

I'm totally confused with those notations and how they work!!

Answer 1

$$\langle x,y\rangle+\langle y,x\rangle=\langle x,y\rangle+\overline{\langle x,y\rangle}=2\text{Re}(\langle x,y\rangle)$$

But for any complex number $z=a+ib\in\Bbb C\;,\;\;a,b\in\Bbb R\;$ , we have that:

$$\text{Re} (z)=a\le\sqrt{a^2+b^2}=|z|$$

Answer 2

If $V$ is a vector space over $F=\mathbb C$ with scalar product $\langle\cdot,\cdot\rangle$, by definition $$\langle y,x\rangle=\overline{\langle x,y\rangle}, $$ for all $x,y\in V$.

Then $$\langle y,x\rangle+\langle x,y\rangle=\overline{\langle x,y\rangle} +\langle x,y\rangle= 2\operatorname{Re}(\langle x,y\rangle).$$

The last equality follows from $$z+\bar{z}=2\operatorname{Re}(z), $$ for all $z\in\mathbb C$.

Answer 3

To first clarify on notation, let $z = a + bi$. If $b=0$, then observe that $z$ is just a real number. When we say $\operatorname{Re}z$ we are referring to the Real part of $z$. So we have $\operatorname{Re}z = a$. And, for its counterpart, when we say $\operatorname{Im}z$ we are referring to the Imaginary part of $z$. So we have $\operatorname{Im}z = b$.

Now, let $z = \langle x, y \rangle = a + bi$. Note that $z$ may be a complex or real depending on $b$. So, $$\langle x,y \rangle + \langle y, x \rangle = \langle x, y \rangle + \overline{\langle x,y \rangle} = z +\bar{z} = a + bi + a - bi = 2a = 2\operatorname{Re}z=2\operatorname{Re}\langle x,y\rangle \quad\Box$$

We see that the notation $\operatorname{Re}z$ allows us to generalize inner products to complex numbers.