Convolution of an $L_{p}(\mathbb{T})$ function $f$ with a term of a summability kernel $\{\phi_n\}$ ... is the result in $L_{p}$?
A remark in my notes says yes but I can't see how to verify it.
As was pointed out to me in a previous question I asked last night, I need to show that the following integral is finite:
$$\int_{-\pi}^{\pi}|\int_{-\pi}^{\pi}f(t-s)\phi_{n}(s)ds|^{p}dt < \infty$$.
One of the properties of a summability kernel is that there exists a $C > 0$ such that $\int_{-\pi}^{\pi}|\phi_{n}(t)|dt\leq C$ for every $n\geq 1$.  I feel like this could help if I could get $\phi$ by itself
 A: The way I usually prove Young's Inequality is using a couple of applications of Hölder's inequality to prove
$$
\left|\int f(x)\;g(x)\;h(x)\;\mathrm{d}x\right|\le\|f\|_u\|g\|_v\|h\|_w\tag{1}
$$
where $\frac1u+\frac1v+\frac1w=1$. Then apply $(1)$ in a tricky way to show
$$
\left|\iint f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\le\|f\|_p\|g\|_q\|h\|_r\tag{2}
$$
where $\frac1p+\frac1q+\frac1r=2$. Taking the supremum of inequality $(2)$ over all $h\in L^r$ such that $\|h\|_r=1$ says that $\|f\ast g\|_s\le\|f\|_p\|g\|_q$ where $\frac{1}{r}+\frac{1}{s}=1$, that is $\frac{1}{s}=\frac{1}{p}+\frac{1}{q}-1$.

Tricky Application of $\mathbf{(1)}$:
Since $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=2$, we have the following 7 relations:
$$
\begin{align}
\left(1-\frac{1}{r}\right) + \left(1-\frac{1}{p}\right) + \left(1-\frac{1}{q}\right) &= 1\\
\left(1-\frac{1}{r}\right) + \left(1-\frac{1}{q}\right) &= \frac{1}{p}\text{ so that }p\left(1-\frac{1}{r}\right) + p\left(1-\frac{1}{q}\right) = 1\\
\left(1-\frac{1}{r}\right) + \left(1-\frac{1}{p}\right) &= \frac{1}{q}\text{ so that }q\left(1-\frac{1}{r}\right) + q\left(1-\frac{1}{p}\right) = 1\\
\left(1-\frac{1}{p}\right) + \left(1-\frac{1}{q}\right) &= \frac{1}{r}\text{ so that }r\left(1-\frac{1}{p}\right) + r\left(1-\frac{1}{q}\right) = 1
\end{align}
$$
Therefore,
$$
\begin{align}
&\left|\iint f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\\
&\le\iint|f(x-y)|\;|g(y)|\;|h(x)|\;\mathrm{d}y\;\mathrm{d}x\\
&={\small\iint\underbrace{|f(x-y)|^{p(1-1/r)}|g(y)|^{q(1-1/r)}}_{\large\text{in }L^w\text{ where }\frac1r+\frac1w=1}\;\underbrace{|g(y)|^{q(1-1/p)}|h(x)|^{r(1-1/p)}}_{\large\text{in }L^u\text{ where }\frac1p+\frac1u=1}\;\underbrace{|f(x-y)|^{p(1-1/q)}|h(x)|^{r(1-1/q)}}_{\large\text{in }L^v\text{ where }\frac1q+\frac1v=1}\;\mathrm{d}y\;\mathrm{d}x}\\
&\le\left(\iint|f(x-y)|^p|g(y)|^q\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/r}\\
&\times\left(\iint|g(y)|^q|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/p}\\
&\times\left(\iint|f(x-y)|^p|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/q}\\
&=\left(\int|f(x)|^p\;\mathrm{d}x\right)^{1/p}\left(\int|g(x)|^q\;\mathrm{d}x\right)^{1/q}\left(\int|h(x)|^r\;\mathrm{d}x\right)^{1/r}
\end{align}
$$
