Let R(T) be the set of all functions $f:\mathbb{R}\mapsto\mathbb{C}$ such that
$f$ is $2\pi$-periodic and integrable over any closed interval of length $2\pi$.
Let $D_{n}$ be the Dirichlet kernel and $K_{n}(f)$ be the Fejer kernel meaning that $D_{n}=\sum_{m=-n}^{n}e_{m}$ , $K_{n}=\frac{1}{n}\sum_{m=0}^{n-1}D_m$ and $e_{m}:\mathbb{R}\mapsto\mathbb{C}$ with $e_{m}(t)=e^{imt},\forall t\in\mathbb{R}$
If $F:\mathbb{R}\mapsto\mathbb{R}$ such that $F=(\sum_{n=1}^{\infty}\frac{|(D_{n}-K_{n})*f|^2}{n})^\frac{1}{2}$ show that $F\in R(T)$ and moreover that $||F||_{2}\leq||f||_{2}$
It's easy to show that $\forall n\in\mathbb{N}$ we have $\phi_{n}:=(D_{n}-K_{n})*f:\mathbb{R}\mapsto\mathbb{C}$ with $\phi_{n}=\sum_{m=-n}^{n}\frac{|m|\widehat{f}(m)}{n+1}e_{m}$.
From this we conclude that $(\phi_{n})_{n\in\mathbb{N}}$ is a sequence of complex trigonometric polynomials of degree at most n thus continuous and $2\pi$-periodic.
Therefore if $\forall n\in\mathbb{N}$ we define $g_{n}:\mathbb{R}\mapsto\mathbb{R}$ with $g_{n}:=\frac{|\phi_{n}|^2}{n}$then the sequence $(g_{n})_{n\in\mathbb{N}}$ is a sequence of non-negative,continuous and $2\pi$-periodic real functions.From the monotone convergence theorem we have that the series $\sum_{n=1}^{\infty}g_{n}$ is Lebesgue integrable over $\mathbb{R}$ thus F is also Lebesgue integrable over $\mathbb{R}$.
First question
From continuity of $g_{n}$ and the monotone convergence we cannot conclude that F is also continuous.What about $g_{n}$?They are not uniformly bounded i think
$||F||_{2}^{2}:=\frac{1}{2\pi}\int_{-\pi}^{\pi}|F(t)|^2dt$
$=\frac{1}{2\pi}\int_{-\pi}^{\pi}lim_{N\rightarrow\infty}\sum_{n=1}^{N}g_{n}(t)dt=lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{n=1}^{N}g_{n}(t)dt$
$=lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{|\phi_{n}(t)|^{2}}{n}dt=lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{n}\frac{1}{2\pi}\int_{-\pi}^{\pi}|\phi_{n}(t)|^{2}dt$
$=lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{n}\frac{1}{2\pi}\int_{-\pi}^{\pi}|\sum_{m=-n}^{n}\frac{|m|\widehat{f}(m)}{n+1}e^{imt}|^{2}dt$
$=lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{n}||\sum_{m=-n}^{n}\frac{|m|\widehat{f}(m)}{n+1}e_{m}||_{2}^{2}$
orthogonality
$=lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{n}\sum_{m=-n}^{n}||\frac{|m|\widehat{f}(m)}{n+1}e_{m}||_{2}^{2}$
$=lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{n}\sum_{m=-n}^{n}\frac{m^{2}|\widehat{f}(m)|^{2}}{(n+1)^{2}}||e_{m}||_{2}^{2}$
$=lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{n}\sum_{m=-n}^{n}\frac{m^{2}|\widehat{f}(m)|^{2}}{(n+1)^{2}}$
Second question
I have tried to write the double sum differently as follows
$\sum_{n=1}^{N}\{[\sum_{n-1\leq|m|\leq n}{m^{2}|\widehat{f}(m)|^{2}}][\sum_{ν=n}^{N}\frac{1}{ν (ν+1)^{2}}]\}$
$\leq\sum_{n=1}^{N}\{[\sum_{n-1\leq|m|\leq n}{n^{2}|\widehat{f}(m)|^{2}}][\sum_{ν=n}^{N}\frac{1}{ν (ν+1)^{2}}]\}$
$=\sum_{n=1}^{N}\{[\sum_{n-1\leq|m|\leq n}{|\widehat{f}(m)|^{2}}][\sum_{ν=n}^{N}\frac{n^{2}}{ν(ν+1)^{2}}]\}$
If i could conclude that $\sum_{ν=n}^{N}\frac{n^{2}}{ν(ν+1)^{2}}\leq 1$ then it would follow that
$||F||_{2}\leq||f||_{2}$
Using the integral test for series convergence and the upper estimates for partial sums of the form
$\sum_{v=n}^{N}h(v)\leq h(n)+\int_{n}^{N}h(t)dt\leq h(n)+\int_{n}^{\infty}h(t)dt$ where $h(t)=\frac{v^2}{t(t+1)^2}\forall t\in[1,\infty )$
we get that $\sum_{ν=n}^{N}\frac{n^{2}}{ν(ν+1)^{2}}\leq 1$