Show that sequnce of functionals $g_n^*(f) = n\displaystyle{\int_0^1 x^nf(x)dx}, f \in C[0,1]$ converges weakly and find its limit functional. Does it converge in the norm of space $C^*[0,1]$?
I don't even know how to start, I guess I have to guess first which is limit functional, and then prove $g_n^*$ converges to it, but I don't have any idea what it could be.
On
The proof can be performed without applying the Weierstrass theorem. Instead we will base on the fact that $C^{2}[0,1]$ functions are dense in $C[0,1].$ Since the norms of the functionals are bounded it suffices to show the weak convergence holds for $C^2[0,1].$ Let $f\in C^2[0,1].$ Then applying the integration by parts two times we get $$g_n^*(f)={n\over n+1}x^{n+1}f(x)\Big\vert_0^1-{n\over n+1}\int\limits_0^1x^{n+1}f'(x)\,dx\\ ={n\over n+1}f(1)-{n\over (n+1)(n+2)}x^{n+2}f'(x)\Big\vert_0^1+{n\over (n+1)(n+2)}\int\limits_0^1x^{n+2}f''(x)\,dx\\ ={n\over n+1}f(1)+O(n^{-1})$$ Therefore $g_n^*(f)\to f(1).$
Regarding weak convergence - you want to prove that for any function $f\in C[0,1]$ the limit exists. One starts with $f$ being of the form $x^k$. It is easy to compute that $g_n(f) = \frac{n}{n+k}$ so the limit is $1$. Hence it is $f(1)$. Then, using linearity of $g_n$ one gets the same for $f(x) = \sum a_kx^k$ i.e. that the limit is $f(1)$. Then, by the fact that the norm of $g_n$ are bounded by 1 and density of polynomials in $C[0,1]$ one gets that the limit exists and is $f(1)$ for any continuous $f$. Lastly, since the limit is $f(1)$ it is just integral against measure with single atom at $1$.