I am reading the first chapter of Y. Katznelson "An introduction to harmonic analysis". There is a definition of homogeneous space $B$ on $\mathbb{T}$, and then it is proved that the trigonometric polynomials form a dense subspace of $B$, and that given $f\in B$ and a summability kernel $(K_n)$, the sequence of convolution products $(K_n\ast f)$ converges to $f$ in the norm of $B$.

What I cannot see is why the trigonometric polynomials or the terms $K_n\ast f$ are contained in $B$.


Does K actually say that the trigonometric polynomials form a dense subspace of $B$? Unless I'm forgetting something about the definition this is not true, for example $B=\{0\}$ is a homogeneous space. What's true is that $TP\cap B$ is a dense subspace of $B$ (where $TP$ is the space of trigonometric polynomials).

Why $K_n*f\in B$ for $f\in B$: First note that by definition $B\subset L^1$, so the convolution exists and lies in $L^1$.

Say $f\in B$. Then $h\mapsto\tau_h f$ defines a continuous map from $\Bbb T$ to $B$. Since the norm on $B$ is translation invariant, $$||\tau_{x+h}f-\tau_x f||=||\tau_h f-f||,$$which says that this map is in fact uniformly continuous. Let's give this map a name: Define $F:\Bbb T\to B$ by $$F(h)=\tau_h f.$$

Now if $\phi$ is continuous then $$\phi*f=\frac1{2\pi}\int_0^{2\pi}\phi(h)F(h)\,dh.$$That's the integral of a continuous $B$-valued function, which hence lies in $B$ (by standard results about vector-valued integrals, or one could say that the proof that the Riemann sums for a complex-valued continuous function converge in $\Bbb C$ shows that the Riemann sums for the integral of a continuous $B$-valued function converge in $B$).

And of course the fact that $B\subset L^1$ shows that if $\phi$ is a trigonometric polynomial then $\phi*f$ is a trigonometric polynomial with coefficients $\hat\phi(n)\hat f(n)$.


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