There is no polynomial $q$ such that $\int_0^1 p(x)q(x)\,dx=p(0)$ for each polynomial $p$. Suppose the inner product on $P(R)$ is defined by $\langle p,q\rangle = \int^1_0 p(x)q(x)dx$.
Let $\phi$ be the linear functional on $P(R)$ defined by $\phi (p) = p(0)$ for each polynomial $p \in P(R)$. Prove that there does not exist $q \in P(R)$ such that $\phi (p) = \langle p, q\rangle$ for every $p \in P(R)$.
This is what I have so far:
Let $e_1,...,e_n$ be an orthonormal basis for $P(R)$.
$\phi (p) = \phi(\langle p, e_1\rangle e_1 + \dots + \langle p, e_n\rangle e_n) = \langle p, e_1\rangle \phi(e_1) + \dots + \langle p, e_n\rangle \phi(e_n)$
Thus, $\phi(e_1)e_1 + \dots + \phi(e_n) e_n = p(0)e_1 + \dots p(0) e_n$
I'm not sure where to go from here, help? Thank you!
 A: Take $p(x)=1$. We get $\int_0^1 q(x)dx=1$. Now take $p(x)=xq(x)$, we get $\int_0^1 x(q(x))^2dx=0$, hence $xq(x)^2=0$ on $[0,1]$ and $q(x)=0$. Hence $\int_0^1 q(x)dx=0=1$.
A: The vector space of all polynomials with real coefficients has infinite dimension, so there is no finite orthonormal basis as stated at the beginning of your work.  What follows the word "Thus" is very mysterious to me, and incorrect.
An alternative to the other suggestions is to consider $p_n(x)=(1-x)^n$ and use the bounded convergence theorem:  If $(f_n)$ is a sequence of integrable functions on $[0,1]$ converging pointwise to an integrable function $f$, and if there is a constant $M>0$ such that $|f_n(x)|<M$ for all $n$ and for all $x$, then $\lim\limits_{n\to\infty}\int_0^1f_n(x)\,dx=\int_0^1f(x)\,dx$.
A: Notice that $\int\limits_{0}^{1}x^{\alpha}a_{n}x^{n}dx$ is a rational function in term of $\alpha$ assuming $\alpha\geq 0$.
Hence $\int\limits_{0}^{1}x^{\alpha}q(x)dx$ is still a rational function in term of $\alpha$.
Remember that rational function is ratio of 2 polynomial. Thus it have only a finite number of root. Hence only a finite number of $\alpha$ exist such that $\int\limits_{0}^{1}x^{\alpha}q(x)dx=0$. But we need an infinite number of them, by plugging in $p(x)=x^{\alpha}$ for $\alpha=1,2,3,\ldots$
A: Hint
Take the restriction $\phi_n$ of $\phi$ on the subspace $\Bbb R_n[x]$ of polynomials with degree less or equal $n$. 
$\phi_n$ is   obviously continuous by (in finite dimensional space all the norms are equivalent)
$$|\phi_n(p)|\le \lVert p\rVert_\infty$$
so by the Riesz's theorem there's a unique $q_0$ such that
$$\phi_n(p)=\langle p,q_0\rangle$$
A: According to the Cauchy-Schwarz inequality we would have $$|p(0)| = |\langle p, q \rangle| \leq \lVert p\rVert\lVert q\rVert$$ for all polynomials $p$. Now consider this inequality for $p(x) = (1-x)^n$ and any $n \in \mathbb{N}$.
A: Consider $x^n$ for $n=0,1,2,\ldots$ and use Weierstrass' theorem.
