Showing Positive Definiteness In Inner Product Let $\mathcal{P}_2$ be the space of all polynomials of degree less or equal to $2$ for all $f,g\in \mathcal{P}_2$ we define:
$$\langle f,g \rangle=\int_0^\infty f(x)g(x)e^{-x} \,dx$$
Prove it is a inner product on $\mathcal{P}_2$

It is easy to show that $\langle f,g \rangle=\langle g,f \rangle$ and $\langle \alpha f+\beta g,h \rangle=\alpha\langle  f,h \rangle+\beta\langle g,h \rangle$
To show positive definiteness we take $\langle f,f \rangle=\int_0^\infty [f(x)]^2 e^{-x} \, dx$ but how can we be sure that $\langle f,f \rangle\geq 0$ and $\langle f,f \rangle=0\iff f=0$?
We know that $e^{-x}&gt; 0$ and $(f(x))^2\geq 0$. Thus, $\displaystyle\int_{0}^{\infty} (f(x))^2 e^{-x}\geq 0$ and this function is the product of continuous functions, therefore, is continuous. Then, $\displaystyle\int_{0}^{\infty} (f(x))^2 e^{-x}=0$ if and only if $(f(x))^2 e^{-x}=0$ if and only if $(f(x))^2=0$ if and only if $f(x)=0$
You also have to show that the expression well-defined. To do this for given $f, g \in \mathcal{P}_2$, choose $M &gt; 0$ such that for all $x \geq M$ we have $\lvert f(x) \cdot g(x)\rvert \leq x^5$ and $x^7 e^{-x}\leq 1$. Then we have
$$
\int_M^\infty\lvert f(x)g(x)e^{-x} \rvert \, dx \leq \int_M^\infty x^5e^{-x} \, dx = \int_M^\infty x^7e^{-x}x^{-2} \, dx \leq \int_M^\infty x^{-2} \, dx &lt; \infty.
$$

It follows that
$$
\int_M^\infty f(x)g(x)e^{-x} \, dx &lt;\infty.
$$