Proving existence and uniquennes of integral Let $\mathcal{C}_0[a,b]$ be the set of continous real valued functions in $[a,b]$, and $\mathcal{P}_0[a,b]$ the subspace of polynomials. Admitting Weierstrass Theorem, show that there exists an unique linear continous functional $J:\mathcal{C}_0[a,b]\rightarrow\mathbb{R}$ such that, for $f(x)=x^n$, $J(f) = \frac{1}{n+1}(b^{n+1}-a^{n+1})$.
My problem is I'm not sure what I can assume from the start, do I assume there exists a continous linear functional with $J(f) = \frac{1}{n+1}(b^{n+1}-a^{n+1})$ and prove it is well defined for other functions, or do I neec to prove that a functionla that satisfies this equality is linear and continous?
 A: Define $J :  C_0[a,b] \to \mathbb{R}$ as $$Jf = \int_a^b f(x)\,dx, \quad\forall f \in C_0[a,b]$$
It is easy to verify that $J$ is linear $J(x^n) = \frac1{n+1}(b^{n+1}-a^{n+1}), \forall n \ge 0$.
Furthermore, $J$ is bounded:
$$|Jf| = \left|\int_a^b f(x)\,dx\right| \le \int_a^b \left|f(x)\right|\,dx \le \int_a^b \left\|f(x)\right\|_\infty\,dx  = (b-a)\|f\|_\infty$$
Now assume that $F :  C_0[a,b] \to \mathbb{R}$ is another bounded linear functional  such that $F(x^n) = \frac1{n+1}(b^{n+1}-a^{n+1}), \forall n \ge 0$. By linearity $J$ and $F$ coincide on all polynomials.
For $f \in C_0[a,b]$, Stone-Weierstrass gives a sequence of polynomials $(p_n)_n$ in $C_0[a,b]$ such that $p_n \to f$ uniformly.
Since $F$ is bounded, we have $Fp_n \to Ff$. On the other hand, we have $Fp_n = Jp_n \to Jf$. By the uniqueness of the limit, we conclude $Ff =Jf$.
Therefore, $J$ is the unique bounded linear functional satisfying $J(x^n) = \frac1{n+1}(b^{n+1}-a^{n+1}), \forall n \ge 0$.
A: I would personally follow these steps:


*

*Define the functional over the polynomials.

*Show it's linear.

*Show it's bounded over the polynomials.

*Use Stone-Weierstrass to conclude that $\mathcal{P}[a, b]$ is dense in $\mathcal{C}_0[a, b]$.

*Conclude that the functional has a unique extension to $\mathcal{C}_0[a,b]$. You can use Hahn-Banach to prove existence if necessary, but since the functional is uniformly continuous, there will always be a unique extension to the completion, $\mathcal{C}_0[a, b]$.

