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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.

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    $\begingroup$ 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
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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,

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

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

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

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