Is there any way to prove that the functional series $$g(t)=\sum\limits_{n=1}^{\infty} b_n\,n\,a\sin(n\,c\,t)$$ is uniform convergent, given $$b_n=\int_{-\pi}^{\pi} f(x)\,\sin(nx)dx$$ $f\in \mathcal{C}^2$, $a,c>0$, and knowing that $\int_{-\pi}^{\pi} f(x)\,\cos(nx)dx=0$
All I could prove is that $\lim\limits_{n\to\infty} b_n\,n=0$, and I am not sure it is correct
You may suppose that $c = 1$ and $a =1$ without lost of generality. Start with what we know from $f $ being of C 2.
Let $T(x) = \frac{1}{2}{a_0} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos (nx) + {b_n}\sin (nx)} \right)}$ be the Fourier series of $f$. Since $f'$ is continuous, $ f $ is absolutely continuous and $T(x)$ converges absolutely and uniformly to $f(x)$ and the differentiated series is the Fourier series of $ f '$, that is
$S(x) = \sum\limits_{n = 1}^\infty {\left( -n{{a_n}\sin (nx) + n{b_n}\cos (nx)} \right)}$
is the Fourier series of $ f '$. Then we can deduce that $T(x)$ converges absolutely and uniformly. More precisely,
$\sum\limits_{k = 1}^\infty {\left| {{a_k}} \right|} < \infty $ and $\sum\limits_{k = 1}^\infty {\left| {{b_k}} \right|} < \infty $ . We shall prove this first then apply this result to the Fourier series of $ f '$.
Since $ f '$ is continuous and so it is square integrable. Let ${B_n} = - n{a_n}$ and ${A_n} = n{b_n}$. Then by the Parseval identity,
$\sum\limits_{n = 1}^\infty {\left( {A_n^2 + B_n^2} \right)} < \infty $.
Therefore,
$\sum\limits_{k = 1}^n {\left| {\frac{{{B_k}}}{k}} \right|} = \sum\limits_{k = 1}^n {\sqrt {\frac{{B_k^2}}{{{k^2}}}} } \le {\left\{ {\left( {\sum\limits_{k = 1}^n {B_k^2} } \right)\left( {\sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} \right\}^{1/2}} \le {\left\{ {\left( {\sum\limits_{k = 1}^\infty {B_k^2} } \right)\left( {\sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}} } \right)} \right\}^{1/2}} < \infty $ .
Hence
$\sum\limits_{k = 1}^\infty {\left| {{a_k}} \right|} < \infty $ . Similarly we can deduce that $\sum\limits_{k = 1}^\infty {\left| {{b_k}} \right|} < \infty $ .
Now we have $f'$ is differentiable and that $f''$ is continuous. By what we just show
$\sum\limits_{k = 1}^\infty {\left| {k{a_k}} \right|} < \infty $ and $\sum\limits_{k = 1}^\infty {\left| {k{b_k}} \right|} < \infty $ .
Now apply Weierstrass M-test, we have $g(t)$ converges absolutely and uniformly.