Young inequality: Generalization on $\mathbb{T}$ space. I'm interested in resolving this question that I find but on the $\mathbb{T}$ space.  (Show that for any $f\in L^1$ and $g \in L^p(\mathbb R)$, $\lVert f ∗ g\rVert_p \leqslant \lVert f\rVert_1\lVert g\rVert_p$.)
Now my question is, is it possible?
I mean I understand quite well the explanation that I found there but, if I take an $f\in L^1(\mathbb{T})$ and a $g\in L^p(\mathbb{T})$ is possible to prove, in the same way, that $f*g\in L^p$ for $1\le p\le \infty$ and that $||f*g||_{p}\le||f||_{1}||g||_{p}$? And if $p=\infty$? Is possible to prove that $||f*g||_{\infty}\le ||f||_{1}||g||_{\infty}$ if $f*g\in\mathbb{C(\mathbb{T})}$? 
Thanks you very much!
 A: Yes the same proof works since it is a finite measure space .
I am giving the same proof here again.
We can define a finite measure $\mu$ on $\mathbb{T}$ in the following way.
$$\mu(B)=\frac 1{\lVert g\rVert_1}\int_B|g(x)|\mathrm dx;$$for any measurable subset set $B$ of $\mathbb{T}$.
We have 
$$\left(\int_{\mathbb T}|f(x-y)|\cdot |g(y)|\mathrm dy\right)^p
=\left(\int_{\mathbb T}|f(x-y)|\mathrm d\mu(y)\right)^p\lVert g\rVert_1^p\leqslant\int_{\mathbb T}|f(x-y)|^p|\mathrm d\mu(y)\lVert g\rVert_1^p.$$
Now integrate over $x$ with respect to Lebesgue measure and use Fubini-Tonelli's theorem to conclude.
$$\lVert f ∗ g\rVert_p^p\leq\int_{\mathbb T}\left(\int_{\mathbb T}|f(x-y)|\cdot |g(y)|\mathrm dy\right)^p\mathrm dx
=\int_{\mathbb T}\lVert g\rVert_1^p\left(\int_{\mathbb T}|f(x-y)|\mathrm d\mu(y)\right)^p\mathrm dx$$$$\leq
\int_{\mathbb T}\lVert g\rVert_1^p\int_{\mathbb T}|f(x-y)|^p|\mathrm d\mu(y)|\mathrm dx=\lVert g\rVert_1^p\int_{\mathbb T}\int_{\mathbb T}|f(x-y)|^p\mathrm dx|\mathrm d\mu(y)|=\lVert g\rVert_1^p\int_{\mathbb T}\lVert f\rVert_p^p|\mathrm d\mu(y)|=\lVert g\rVert_1^p\lVert f\rVert_p^p\int_{\mathbb T}|\mathrm d\mu(y)|=\lVert g\rVert_1^p\lVert f\rVert_p^p\mu(\mathbb T).$$
And it is clear from the definition of $\mu$ that
$$ \mu(\mathbb T)=\frac 1{\lVert g\rVert_1}\int_{\mathbb T}|g(x)|\mathrm dx=1$$
for $p=\infty$
$$\lvert f ∗ g(x)\rvert\leq\int_{\mathbb T}|f(x-y)|\cdot |g(y)|\mathrm dy
\leq\int_{\mathbb T}\lVert f\rVert_{\infty}\cdot |g(y)|\mathrm dy\leq\lVert f\rVert_{\infty}\lVert g\rVert_1$$
$$
\implies \lVert f ∗ g\rVert_{\infty}\leq\lVert f\rVert_{\infty}\lVert g\rVert_1
$$
