# Modified inner product

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 $a<b$?

• $f$ and $g$ are orthogonal with respect to which inner product? – marco trevi Jan 7 '15 at 20:14
• @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? – Sidd Jan 7 '15 at 20:16
• 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. – whacka Jan 7 '15 at 20:25
• the converse is true: orthogonality implies linear independence. – marco trevi Jan 7 '15 at 20:33

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 $b=-a$.