= Solution
A <summability kernel on the circle> is a sequence of continuous periodic functions with
$$
\int k_n\,dm=1,\qquad \sup_n\|k_n\|_1=C<\infty,\qquad\int_{\delta\leq|t|\leq\pi}|k_n(t)|\,dm(t)\longrightarrow0\quad(0<\delta<\pi).
$$
The frequently used stronger condition of uniform convergence to zero off each neighborhood of zero implies the last condition. For $f$ in a homogeneous <Banach space>,
$$
\|f*k_n-f\|_B\leq\int|k_n(s)|\|\tau_sf-f\|_B\,dm(s).
$$
Given $\varepsilon>0$, translation continuity bounds the difference inside a sufficiently small neighborhood by $\varepsilon$; its integral is at most $C\varepsilon$. Outside, use $2\|f\|_B$ and the vanishing tail condition. Hence $f*k_n\to f$ in $B$.
With the indexing needed in part (c), the <Fejér kernel> is
$$
K_n(t)=\frac1{n+1}\left|\sum_{r=0}^n e^{irt}\right|^2=\sum_{|j|\leq n}\left(1-\frac{|j|}{n+1}\right)e^{ijt},\qquad n\geq0.
$$
In particular $K_0=1$. The squared-modulus expression proves $K_n\geq0$, and its constant <Fourier coefficient> proves $\int K_n\,dm=\|K_n\|_1=1$. The <geometric series> yields
$$
K_n(t)=\frac1{n+1}\frac{\sin^2((n+1)t/2)}{\sin^2(t/2)}\leq\frac1{(n+1)\sin^2(\delta/2)}\quad(\delta\leq|t|\leq\pi).
$$
Thus these are summability kernels. For $T_f:g\mapsto f*g$, part (a) gives $\|T_f\|\leq\|f\|_B$. Since $\|K_n\|_1=1$ and $T_fK_n\to f$ in $B$, the reverse bound follows:
$$
\boxed{\|T_f:L^1(\mathbb T)\to B\|=\|f\|_B.}
$$
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