Let $V$ be the vector space of all continuous functions $f\colon [-1,1]\to\Bbb R$. Prove that $(f, g) = \int_{-1}^{1} f(t)g(t) \,\mathrm dt$ defines an inner product.
I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this
I understand that for $(f, g) = \int_{-1}^{1} f(t)g(t) \,\mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.
Where i have; $(x,y) = (y,x)$ for all $x,y \in V$
and
$(x,x) \geq 0$ for all $x \in V$ and $(x,x)=0$ only if $x$ is the zero vector
I assumed it would simply be plugging in but im not entirely sure.
We have $$(f,g)=\int_{-1}^{1} f(t)g(t) dt=\int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$. And we have $$(f,f)=\int_{-1}^{1} (f(t))^2 dt\geq 0$$ for every f by $(f(t))^2\geq 0$ and the monotony of the integral. We now have to show that $(f,f)=0$ implies $f=0$. It is standard lemma that for a continuous function $g\geq 0$ we have $$\int_{a}^{b} g(t) dt=0\Rightarrow g=0.$$ We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2\geq 0$ is clear) and get $$0=(f,f)=\int_{-1}^{1} (f(t))^2 dt\Rightarrow f^2=0\Rightarrow f=0.$$ If you don't know the lemma, I can give you a prove for it.