inner product on a space P

Question

Prove that $$ \langle f, g\rangle = \int_{-1}^1 f(x)g(x)\,dx $$ is an inner product on the space $P_n$.

My partial solution:

We prove the axioms:

1- Symmetric property: $$ \langle f, g \rangle = \int_{-1}^1 f(x)g(x)\,dx = \int_{-1}^1 g(x)f(x)\,dx = \langle g, f \rangle $$

2- Linearity property: $$\langle af+g,h \rangle \int_{-1}^{1} \bigl(a f(x)+g(x)\bigl) h(x) \, dx = a\int_{-1}^{1} f(x)h(x) \, dx + \int_{-1}^{1} g(x)h(x) \, dx=a\langle f,h\rangle +\langle g,h\rangle $$

3- Positive definite property: $$\langle f, f\rangle = \int_{-1}^1 f(x)^2 \, dx \ge 0$$ and $$ \langle f, f\rangle = 0 $$ if and only if $f$ is the constant $0$ function.

I'm stuck at the third axiom. I gave it a few tries but I feel lost. any help would be great.

Answer 1

For the third axiom,

$$(\forall x\in[-1,1])\;$$ $$ f^2(x)\ge 0\implies \int_{-1}^1f^2(x)dx\ge 0$$

Assume now that $$\int_{-1}^1f^2=0\text{ and } (\exists c\in (-1,1))\;:\; f^2(c)>0$$

by continuity of $ f^2 $, there exists $\eta>0 $ small such that

$$(\forall x\in(c-\eta,c+\eta))\;\; $$ $$f^2(x)\in(f^2(c)-\frac{f^2(c)}{2},f^2(c)+\frac{f^2(c)}{2})$$

$$\implies (\forall x\in(c-\eta,c+\eta))$$ $$\; f^2(x)>\frac{f^2(c)}{2}>0$$

$$\implies \int_{-1}^1f^2=\int_{-1}^{c-\eta}f^2+\int_{c-\eta}^{c+\eta}f^2+\int_{c+\eta}^1f^2$$ $$\ge \int_{c-\eta}^{c+\eta}f^2>0$$

which contradicts the premisse $\int_{-1}^1f^2=0$.

so, $\forall c\in(-1,1) f(c)=0$.

the same when $ c=\pm 1$.

Answer 2

Suppose $<f,f>=0$.

Since $f$ is a polynomial, $f$ is continuous. Therefore, $f^2$ is positive and continuous.

Let’s note $F(t)=\int_{-1}^t f^2(x)dx$, such that $F’(t)=f^2(t), \forall t \in [-1,1]$.

Since $F’(t)=f^2(t)\geq 0$, $F(t)$ is increasing on $[-1,1]$. But, $F(-1)=F(1)=0$. Therefore, $F(t)=0, \forall t \in [-1,1]$.

Hence, $F’(t)=f^2(t)=0$. Which means that $\forall t \in [-1,1]$, you have $f(t)=0$. So $f=0$.