Inner Product Spaces and Minimum Polynomials 
Problem
Let $V$ be the vector space of real polynomials $\mathbb{R}[x]$ endowed with the inner product
$$ (f,g)=\int_{-\infty}^{\infty}e^{-|x|}f(x)g(x)dx $$
By considering the series of subspaces $\{V_n\}$ where $V_n=\{f(x) \in \mathbb{R}[X]: \deg(f) \leq n\}$ or otherwise, show that there exist unique monic polynomials $\varphi_n(x)$ for $n \geq 0$ such that 
$$\int_{-\infty}^{\infty}e^{-|x|}\varphi_n(x)g(x)dx=0$$ whenever deg
  $g<n$, and find $\varphi_n(x)$ for $n=0,1,2$.

Progress
We look to identify $(V_n)^{\perp}$ for  each $n$, i.e.
$$(V_n)^{\perp}=\{f(x) \in \mathbb{R}[X]: \langle f,g \rangle=0 \text{ for all } g \in V_n\}$$
If we can show this has dimension $1$, then we can show the existence of a unique monic polynomial. Any thoughts on how this is done (or an alternative method for the problem) would be greatly appreciated. Regards.
 A: This problem should remind you of Gram-Schmidt, but if you are interested in showing only the existence (and uniqueness) of the $\varphi_n$, and not interested in the final formula, then one can stop short of all the calculations. That said, I recommend that you are not as lazy as I am; the calculations might end up illuminating. :) 
All one needs to understand is the abstract operation of orthogonal projection. Consider a vector space $V$ and a finite dimensional subspace $E \subseteq V$. Then the orthogonal projection of $v$ onto $E$ is the unique vector $w \in E$ such that $v - w \perp E$. It is a standard fact that such a projection always exists and is unique; let us denote this $\operatorname{Proj}(v, E)$. 
If you know this fact (or take it for granted), then this problem is easy: 
$$
\varphi_n(x) = x^n - \operatorname{Proj}(x^n, V_{n-1})
$$
is the unique monic polynomial of degree $n$ such that $(\varphi_n, g) = 0$ for all $g \in V_{n-1}$. (Note that the projection, being in $V_{n-1}$, is a polynomial of degree $n-1$.) Now, it is an altogether a different matter how to evaluate the projection term. It does not come out to be a nice expression, and moreover, it's not at all clear to me that it is intended here. 

To complete the picture, let us also see why the projection as defined always exists and is unique. 
Uniqueness. Suppose there exist $w_1, w_2 \in E$ such that $v - w_1 \perp E$ and $v - w_2 \perp E$. Then $w_1 - w_2 = (v - w_2) - (v- w_1)$ is also orthogonal to $E$. In particular, since $w_1 - w_2$ itself is in $E$, it is orthogonal to itself; that is, $w_1 - w_2 = 0$. Thus the projection -- when it exists -- is unique.
Existence. Since $E$ is a finite dimensional real* inner product vector space, it possesses an orthonormal basis. Let $u_1, u_2, \ldots, u_r$ be such an orthonormal basis, where $r = \dim E$. Then the projection of a given vector  can simply be written down as:
$$
w = \operatorname{Proj}(v, E) := \sum_{i=1}^r (v, u_i) u_i.
$$
It is easy to see that $v-w$ is orthogonal to each basis vector $u_i$, and hence is orthogonal to the whole subspace $E$. 
Of course, if, in a concrete setting, we want to find the projection, we should compute an orthonormal basis for $E$. This is usually done using the Gram-Schmidt process, as was alluded to in the comments. 

*I assume real base field; the complex case is not too different. 
