# Gram-Schmidt orthogonalization in $L^2$ proof

Hi I am trying to understand a claim made in my textbook and was wondering if anyone can show formally that the set of functions $g_n$ is indeed an orthogonal set.

The claim:

Let $f_1, f_2, f_3, \dots$ be an independent set of functions in $L^2$. Define the set of functions $g_n$ by

$g_1 = f_1$,

$g_2 = f_2 - \frac{(f_2, g_1)}{||g_1||^2}g_1$,

$g_3 = f_3 - \frac{(f_3, g_2)}{||g_2||^2}g_2 - \frac{(f_3, g_1)}{||g_1||^2}g_1$,

What would be the way to show mathematically that $g_n$ is an orthogonal set?

Proceed by induction.

First show that $g_2 \bot g_1$ by evaluating $\langle g_{2}, g_1 \rangle$ and showing that it is zero.

$\langle g_{2}, g_1 \rangle = \langle f_{2} - \langle f_{2}, g_1 \rangle {g_1 \over \|g_1\|^2}, g_1 \rangle =\langle f_{2}, g_1 \rangle - \langle f_{2}, g_1 \rangle { \langle g_{1}, g_1 \rangle \over \|g_1\|^2} = 0$.

Now suppose that the $g_1,...,g_n$ are orthogonal.

Now show that $g_{n+1} \bot g_k$ for $k=1,...,n$ by computing $\langle g_{n+1}, g_k \rangle$ and showing that it is zero.

In order to show that the $g_k$ are non zero, you need to use independence of the $f_k$.

Again, we use induction. We see that $\operatorname{sp} \{ g_1 \} = \operatorname{sp} \{ f_1 \}$.

Now suppose $\operatorname{sp} \{ g_1,...,g_n \} = \operatorname{sp} \{ f_1 , ..., f_n\}$. Then the formula for $g_{n+1}$ shows that if $g_{n+1} = 0$, we would have $f_{n+1} \in \operatorname{sp} \{ f_1 , ..., f_n\}$, which would contradict the $f_k$ being independent.

Since $g_{n+1} \in \operatorname{sp} \{ g_1,...,g_n, f_{n+1} \}$, we see that $g_{n+1} = \alpha f_{n+1} + \sum_k \alpha_k g_k$ and $\alpha \neq 0$, hence $f_{n+1} = {1 \over \alpha} ( g_{n+1} - \sum_k \alpha_k g_k )$, so we see that $\operatorname{sp} \{ g_1,...,g_n, f_{n+1} \} = \operatorname{sp} \{ g_1,...,g_n, g_{n+1} \} = \operatorname{sp} \{ f_1 , ..., f_n, f_{n+1}\}$.

• because $f_1, f_2, \dots$ are an independent set of functions in $L^2$ it is true that $(f_i, f_j) = 0$ for $i \neq j$ right? – Lucile Ackley Nov 19 '16 at 20:46
• No. Independence just means that if $\sum_n \alpha_k f_k = 0$ (for any finite collection) then $\alpha_k = 0$. – copper.hat Nov 19 '16 at 20:48
• ok so I'm just a bit confused on the algebra of the base case. I have $(g_1, g_2) = (f_1, f_2 - \frac{(f_1,g_1)}{||g_1||^2}g_1) = (f_1, f_2 - \frac{(f_2, f_1)}{||f_1||^2}f_1)$ can you explain why this is 0? – Lucile Ackley Nov 19 '16 at 20:50
• Keep expanding it, $\langle a, b-c \rangle = \langle a, b \rangle - \langle a, c \rangle$ and $\langle a, \alpha b \rangle = \alpha \langle a, b \rangle$. – copper.hat Nov 19 '16 at 20:51
• I added an elaboration above. Note that $\langle h,h \rangle = \|h\|^2$. – copper.hat Nov 19 '16 at 21:02