Questions in Orthogonal polynomials and Integration

Question

$Q_i(x) (i=1,n)$ is orthogonal with respect to $w(x)$ such that $(a \le x \le b,w(x)\ge0)$.

$ N_i =\int_a^b Q_i^2(x)w(x)dx$

Since $w(x) \ge 0 $ , then it follows that $N_i \gt 0 $.

I don't understand the sequence that "Since $w(x) \ge 0 $ , then it follows that $N_i \gt 0 $."

Can you help me? Thanks.

Answer 1

It is known that $|\int_{a}^{b} f(x) dx | \leq \int_a^{b} |f(x)| dx$. So $|\int_{a}^{b} Q_i^2(x) w(x) dx| \leq \int_{a}^{b} |Q_i^2(x) w(x)| dx = \int_{a}^{b} Q_i^2(x) w(x) dx$. Hence $\int_{a}^{b} Q_i^2(x) w(x) dx \geq 0$.

Answer 2

As a square of real number, $Q_i^2(x) \geq 0, \,\, \forall a \leq x \leq b$.

Thus $w(x) \ge 0 , \,\, \forall a \leq x \leq b$ implies $Q_i^2(x)w(x) \ge 0 , \,\, \forall a \leq x \leq b.$

Hence $N_i$, as the integration of a non-negative function over $[a,b]$, is non-negative. $$ N_i =\int_a^b Q_i^2(x)w(x)dx \geq 0.$$