Show that the sum of a function series is bounded

(ANSWER)

Theorem. (Konard Knopp. "Theory and Applications of Infinite Series".Dover.1990.pg.348) The series $\sum a_{\nu}(t)b_{\nu}(t)$ is uniformly convergent in $J$ if the series $\sum |b_{\nu}-b_{\nu+1}|$ converges uniformly in $J$, and the series $\sum a_{\nu}$ has uniformly bounded partial sums, provided the functions $b_{\nu}(t)\rightarrow 0$ uniformly in $J$.

Let $\delta$ be any fixed number of $(0,1/2)$. Fix also $x\in(-\pi,\pi)$. Set $$ b_{\nu}=\frac{1}{\nu^{1/2+\delta}}e^{-\nu^2t} $$ and $$ a_{\nu}=\frac{2(-1)^{\nu+1}}{\nu^{1/2-\delta}}\sin(\nu x). $$ Then clearly for $t\geq 0$ $$ \sum^{\infty}_{\nu=1}\left|\frac{e^{-\nu^2 t}}{\nu^{1/2+\delta}}-\frac{e^{-(\nu+1)^2t}}{(\nu+1)^{1/2+\delta}}\right|=e^{-t}<\infty(uniformly). $$ and $b_{\nu}(t)\rightarrow 0$. Also if $x=y-\pi$, then $$ \sum_{1\leq\nu\leq M}\frac{2(-1)^{\nu+1}}{\nu^{1/2-\delta}}\sin(\nu x)=-2\sum_{1\leq\nu\leq M}\frac{1}{\nu^{1/2-\delta}}\sin(\nu y)\tag 1 $$ is uniformly bouded in $\textbf{R}$. This last argument follows from the well known fact that if $c_{\nu}$ is null and monotone then $\sum c_{\nu}\sin(\nu y)$ is uniformly convergent everywhere in $\epsilon\leq y\leq 2\pi-\epsilon$, $0<\epsilon<\pi$ Or equivalently $\epsilon-\pi\leq x\leq \pi-\epsilon$ (see the above reference pg.349). The cace $x=\pm\pi$ is trivial. QED