Discrete measures of orthogonal polynomials Disclaimer this is a hw problem I DO NOT want a full answer I just want to know if I am on the right track. If d$\lambda(t)=$d$\lambda_N(t)$ is a discrete measure with exactly $N$ support points $t_1, t_2,...,t_N$, and $\pi_j(t)=\pi_j(\cdot,\text{d}\lambda_N)$, $j=0,1,...,N-1$, are the corresponding (monic) orthogonal polynomials, 
let $\pi_N(t)=(t-\alpha_{N-1})\pi_{N-1}(t)-\beta_{N-1} \pi_{N-2}(t)$, with $\alpha_{N-1}$, $\beta_{N-1}$ defined as before. Show that $\pi_N(t_j)=0$ for $j=1,2,...,N$. 
So we have 
$$
\alpha_k=\frac{(t \pi_k, \pi_k)}{(\pi_k,\pi_k)} \qquad \beta_k=\frac{(\pi_k,\pi_k)}{(\pi_{k-1},\pi_{k-1})}
$$
So
$$
\pi_N(t)=\left(t- \frac{(t \pi_{N-1}, \pi_{N-1})}{(\pi_{N-1},\pi_{N-1})}   \right) \pi_{N-1}(t)- \left(   \frac{(\pi_{N-1},\pi_{N-1})}{(\pi_{N-2},\pi_{N-2})} \right) \pi_{N-2}(t)
$$
Now we take $t=t_j$ for some $j=1,2,...,N$
$$\pi_N(t_j)=\left(t_j- t_j\frac{(\pi_{N-1}, \pi_{N-1})}{(\pi_{N-1},\pi_{N-1})}   \right) \pi_{N-1}(t_j)- \left(   \frac{(\pi_{N-1},\pi_{N-1})}{(\pi_{N-2},\pi_{N-2})} \right) \pi_{N-2}(t_j)
$$
We can take out the $t_j$ of the inner product since $t_j \in \mathbb{R}$ then we have
\begin{align*}
\pi_N(t_j)&=\left(t_j- t_j \cdot 1   \right) \pi_{N-1}(t_j)- \left(   \frac{(\pi_{N-1},\pi_{N-1})}{(\pi_{N-2},\pi_{N-2})} \right) \pi_{N-2}(t_j) \\
&=- \left(   \frac{(\pi_{N-1},\pi_{N-1})}{(\pi_{N-2},\pi_{N-2})} \right) \pi_{N-2}(t_j) \\
\end{align*}
Then since $(\pi_{N-1},\pi_{N-1})=(t\pi_{N-2},\pi_{N-1})$ then maybe this innerproduct must be zero?
This seams like this is not right My question is am I on the right track and if note where did I go wrong.
 A: Definitely on the wrong track. When you set $t=t_j$ you can't set $\left(t\pi_{N-1},\pi_{N-1}\right)=\left(t_j\pi_{N-1},\pi_{N-1}\right)$ because the $t$ outside the inner product and the $t$ inside are different things. It's sort of bad notation to use the same symbol for different things and if you fixed this up you would not make this mistake.  
I can see the solution, but you said you didn't want a spoiler, so...
EDIT: We know that $\pi_N(t)$ exists because you constructed it via the Gram-Schmidt process in your original question. Suppose there is another monic polynomial $\sigma_N(t)$ of degree $N$ such that $\left(\sigma_N,\rho_{N-1}\right)=0$ for any polynomial $\rho_{N-1}(t)$ of degree not exceeding $N-1$. Then $\pi_N(t)-\sigma_N(t)$ is a polynomial of degree not exceeding $N-1$ so
$$\left(\pi_N-\sigma_N,\pi_N-\sigma_N\right)=\left(\pi_N,\pi_N-\sigma_N\right)-\left(\sigma_N,\pi_N-\sigma_N\right)=0-0=0$$
However,
$$\left|\pi_N(t)-\sigma_N(t)\right|^2\ge0$$
So it must be that
$$\pi_N(t_j)-\sigma_N(t_j)=0$$
For each support point $t_j$. A polynomial of degree at most $N-1$ with $N$ zeros is the zero polynomial, so $\pi_N(t)=\sigma_N(t)$ Having proved uniqueness we construct the unique
$$\pi_N(t)=\prod_{j=1}^N\left(t-t_j\right)$$ and observe that it's orthogonal to all polynomials of lesser degree since it's zero at all the support points. Having proved uniqueness, we know that it's the only one and it does in fact satisfy the conclusion of the theorem.
