 Solution

Use $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and normalized Haar measure $dm=dsdt/(2\pi {)}^2$ on the torus. Here $|s|$ means distance to zero modulo $2\pi$, represented in $[-\pi ,\pi ]$; an ordinary coordinate in $[0,2\pi )$ would not express concentration at the origin correctly.

Put $B(s,t)=2+\cos s+\cos t$, $I_N=\int B^{N}dm$, and $A_N=I_N^{-1}$. Since $0\le B\le 4$ and $B(0,0)=4$, $I_N>0$, the kernel $K_N=A_N{B}^N$ is nonnegative, and $\int K_{N}dm=1$ exactly. Fix $0<\delta \le \pi$. On the set where either $|s|\ge \delta$ or $|t|\ge \delta$, $B\le q=3+\cos \delta <4$. Choose $q^{\prime }$ with $q<q^{\prime }<4$. continuity provides a neighborhood $U$ of the origin of positive measure $v$ where $B\ge q^{\prime }$. ConsequentlyFor $0<\epsilon \le \pi$, take $\delta =\epsilon$ and then $N$ large enough that this bound is at most $\epsilon$. For $\epsilon >\pi$ the exterior conditions are empty. This proves all the requested kernel properties, with $\boxed{A={\left[\int (2+\cos s+\cos t{)}^{N}dm\right]}^{-1}}$.

Now let $M=\Vert f{\Vert }_{\infty }$ and define $P_N=K_N\ast f$. The powered-cosine approximate identity on a torus is a trigonometric polynomial, since its base is a finite sum of complex exponentials and its exponent is an integer. Expanding $K_N$ and integrating its finitely many Fourier terms shows that $P_N$ is also a trigonometric polynomial; it is real-valued because both factors are real. Translation invariance and normalization giveGiven $\eta >0$, uniform continuity of $f$ supplies $\delta >0$ making the difference in brackets smaller than $\eta /2$ on the square $|s|,|t|<\delta$, uniformly in $(x,y)$. The contribution from that square is at most $\eta /2$. On its complement it is at most $2M{\sup }_{\max (|s|,|t|)\ge \delta }{K}_N$, which tends to zero by the preceding estimate. Therefore $\boxed{\Vert P_N-f{\Vert }_{\infty }\to 0}$, proving uniform approximation without assuming any Fourier-series convergence of $f$.