The definition, provided in Baby Rudin, for an orthogonal system of functions on $[a,b]$, is the following

$\textbf{8.10 Definition}$ Let $\{\phi(n)\}_{n \in \mathbb{N}}$ be a sequence of complex functions on $[a,b]$, such that $$\int_a^b \phi_n(x)\overline{\phi_m(x)}dx = 0 \qquad (n \neq m).$$ Then $\{\phi_n\}$ is said to be an orthogonal system of functions on $[a,b]$. ...

I am wondering why the conjugate is needed in the second function in the integral, i.e., $\overline{\phi_m(x)}$, and how to understand the meaning behind? Why is orthogonality defined in such way?

I don't have a formal education in complex analysis, so I will greatly appreciate it if you can explain in a undergrad level before complex analysis.

  • 2
    $\begingroup$ The reason you need that is that you want the norm of a function, the inner product of the function with itself, to be a real, positive number. Without that complex conjugate, $\int (\phi(x))^2 dx$, is a non-real number. With it, $\int \phi(x)\overline{\phi(x)} dx$ is a positive real number. $\endgroup$ – user247327 Apr 25 '18 at 20:18
  • $\begingroup$ @user247327 Thanks it helped! But then how can we assume that {phi}n and {phi}m are orthogonal where n!=m by proving that {phi}n and bar{{phi}m} are orthogonal? $\endgroup$ – KYHSGeekCode Aug 13 '18 at 10:19

To add some details of the proof which was originally suggested by @user247327, I'll publish that to answer.

Like vectors we can define the term "Orthogonal" be the function dot product of the functions $f$ and $g$ ($<f, g>$) be $0$.

And when $f=g$, $<f, g>=<f, f>=||f||$, is the norm of the function.

So the definition of your book is that functions $f$ and $g$ are orthogonal when $<f, g>=0$.

So your question can be interpreted to:

Why is $<f, g>$ is defined as $\int f(x) \overline{g(x)} dx$ ?

That's because we want $a^2+b^2$, not ${(a+bi)}^2$ to get the norm of a complex number $a+bi$. We use $z \bar z$ instead of $z^2$.

Then how can we know that $<f, g>=0$ when $<f, \overline g>=0$?

You can prove that easily by iterating $f(x) \overline{g(x)}$ by $x$.

Let's iterate $x$ dispersely to be intuitive.

So let $x_i$ be $i$th value of $x$, and if we think that similarly, we get:

$$ <f, g>= \lim_{n \to \infty} { \sum_i^n {f(x_i) \overline{g(x_i)}}\frac 1n}$$

And if we let $$f(x_i)=a_i+b_i i, g(x_i)=c_i+d_i i\\ \text{where}\\ a, b, c, d \in \mathbb R$$,

it will be fine if we prove that

$(a_i+b_i i)(c_i + d_i i)=0$ when $ (a_i+b_i i)(c_i-d_i i) =0$.

So to finish,

$$(a_i c_i + b_i d_i) + (b_ic_i - a_id_i)i =0$$

$$\therefore a_ic_i + b_id_i =0, b_ic_i - a_id_i=0$$

$$\therefore (a_i+b_i i)(c_i-d_i i) = (a_ic_i + b_id_i) + (b_ic_i-a_id_i)i=0$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.