0
$\begingroup$

Given two real valued orthogonal functions, say $f(x)$ and $g(x)$, if we define an inner product $$ \langle f,g\rangle \ = \ \int_a^b f(x) g(x) dx,$$

which we know satisfies the properties of an inner product, namely $\bf positive \ definiteness$, $\bf linearity$ in the first argument, and $\bf conjugate \ symmetry$.

Is it true that $\langle f,g\rangle = 0$ for any values of $a$ and $b$ such that $a<b$?

$\endgroup$
  • $\begingroup$ $f$ and $g$ are orthogonal with respect to which inner product? $\endgroup$ – marco trevi Jan 7 '15 at 20:14
  • $\begingroup$ @marcotrevi: Isn't orthogonality universally defined? I mean, given that two functions are linearly independent, can we not deduce that their inner product must also be zero? $\endgroup$ – Sidd Jan 7 '15 at 20:16
  • 2
    $\begingroup$ You said we define an inner product. Singular, not plural. But $\langle\cdot,\cdot\rangle$ depends on $a$ and $b$, which you seem to be allowing to vary. Which means a different inner product for each choice of $a,b$. So what's going on? Which inner product are $f,g$ orthogonal wrt? (Also conjugate symmetry isn't relevant to a real vector space, only a complex vector space.) And no, linearly independent vectors can easily, easily fail to be orthogonal. Just look at $\Bbb R^2$ to see how wrong that idea is: there are way more linearly independent pairs of vectors than orthogonal pairs. $\endgroup$ – whacka Jan 7 '15 at 20:25
  • 1
    $\begingroup$ the converse is true: orthogonality implies linear independence. $\endgroup$ – marco trevi Jan 7 '15 at 20:33
0
$\begingroup$

Usually not: the choice of $a$ and $b$ is crucial. For instance, consider the orthogonal family of exponential functions $e^{inx}$. Two of these are orthogonal, that is, $$ \langle e^{inx},e^{imx}\rangle = \int_a^b e^{i(n-m)x}\,dx=0\quad(m\neq n) $$ if and only $b-a$ is a multiple of $2\pi$.

Edit: I see now that you asked for real-valued functions. Similar examples work; I chose the exponential functions because they are simple. For instance,the functions $x^n$ and $x^m$, if $n$ and $m$ are nonnegative and $n+m$ is odd, are orthogonal if and only if $b=-a$.

$\endgroup$

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.