# Orthogonality of Two Functions

1. Homework Statement

Show that:

$$\varphi_{0}(x) = f_{0}(x)$$

and

$$\varphi_{1}(x) = f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)$$

are orthogonal on the interval [a,b].

2. Homework Equations

Orthogonal functions satisfy:

$$\left\langle\right\varphi_{m},\varphi_{n}\rangle = \int^{b}_{a}\varphi_{m}(x)\varphi_{n}(x)dx = g(m)\delta_{mn}$$

Where, $$\delta_{mn}$$ is the Delta Kronecker.

Also:

$$\left\langle\right\varphi_{m},\varphi_{m}\rangle = \left\|\varphi_{m}\right\|^{2}$$

3. The Attempt at a Solution

Since m and n (0 and 1) are not equal, the Delta Kronecker is zero and therefore the proof is a matter of proving that:

$$\left\langle\right\varphi_{0},\varphi_{1}\rangle = 0$$

Having substituted the functions into the inner product formula in 2:

$$\left\langle\right\varphi_{0},\varphi_{1}\rangle = \int^{b}_{a}\varphi_{0}(x)\varphi_{1}(x)dx = \int^{b}_{a}f_{0}(x)\left[f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)\right]dx$$

Because the Delta Kronecker is zero, all I have to do is show that:

$$\int^{b}_{a}f_{0}(x)\left[f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)\right]dx = 0$$

I'm unsure as to whether I should use integration by parts to do the resulting integral because there is another integral embedded in the $$\varphi_{1}(x)$$ function; which (because it is a definite integral) would be tricky to differentiate or integrate.

You really don't need to bring the Kronecker delta stuff into the picture. It is completely superfluous. You do need to show that $\langle \varphi_0, \varphi_1 \rangle = 0$ as that is the definition of orthogonality.
Why integrate by parts? Use $\langle f, g \rangle \equiv \int_a^b f(x)g(x)\,dx$ and $\varphi_0(x) = f_0(x)$