Let $s$ $$ \sum_{i,j=0}^{n}s_{i+j}c_ic_j\ge 0$$ for all $c_i$. I know that if the scalars $c_i$'s are zero then the quadratic form $$ \sum_{i,j=0}^{n}s_{i+j}c_ic_j = 0.$$ How can I justify that if $$ \sum_{i,j=0}^{n}s_{i+j}c_ic_j = 0.$$ then $c_1=c_2= \cdots c_n=0$ (Note we assume that the $s_{i+j}$ are non-zero). I will be extremely happy if someone can please help me with this.
I tried expanding the whole quadratic form but that doesn't seem to help me. Thanks.
Unfortunately this isn't true as written.
Consider the quadratic form $s_0c_0^2 + 2s_1c_0c_1 + s_2c_1^2$. Take $s_0 = s_1 = s_2 = 1$. Then the form is $(c_0 + c_1)^2$. This has the nontrivial zero $c_0 = -c1$.
So we have a nontrivial quadratic form which is always nonnegative, but it has infinitely many nontrivial zeros.
Let's clarify a couple things.
The sequence $(s_n)$ is assumed to be positive definite. That translates to $\sum\limits_{i,j=0}^{n}s_{i+j}c_i\bar{c_j} > 0$ for nonzero $c$. The strict $>$ and nonzero $c$ are really important here. First of all it implies that $s_0$ is nonzero, and it makes sense to talk about normalizing the sequence $s$ no matter what.
Second of all it implies that, for your putative norm $||P||^2 = \sum\limits_{i,j=0}^{n}s_{i+j}c_i\bar{c_j}$ (where $P = \sum\limits_{i=0}^{n/2}c_ix^i)$, you have that
$||P||^2 = 0$ iff $c = 0$, which is I think the problem you were having.
You cannot. E.g. when all $s_k$s are zero, the sum is always zero regardless of the values of the $c_j$s.
In general, what you want to prove is correct if the associated $(n+1)\times(n+1)$ Hankel matrix $H=(h_{ij})_{0\le i,j\le n}$ given by $h_{ij}=s_{i+j}$ is positive definite. See Positive definite sequence and its corresponding determinant. for more details.
Please look at this image. You will see that I need to show the asked question for $<,>$ to be an inner product. Checking that $