Convolution of an $L_{p}(\mathbb{T})$ function $f$ with a term of a summability kernel $\{\phi_n\}$
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, $$ \begin{align} &\left|\iint f(x-y)\;g(y)\;h(x)\;\mathrm{d}y\;\mathrm{d}x\right|\\ &\le\iint|f(x-y)|\;|g(y)|\;|h(x)|\;\mathrm{d}y\;\mathrm{d}x\\ &={\small\iint\underbrace{|f(x-y)|^{p(1-1/r)}|g(y)|^{q(1-1/r)}}_{\large\text{in }L^w\text{ where }\frac1r+\frac1w=1}\;\underbrace{|g(y)|^{q(1-1/p)}|h(x)|^{r(1-1/p)}}_{\large\text{in }L^u\text{ where }\frac1p+\frac1u=1}\;\underbrace{|f(x-y)|^{p(1-1/q)}|h(x)|^{r(1-1/q)}}_{\large\text{in }L^v\text{ where }\frac1q+\frac1v=1}\;\mathrm{d}y\;\mathrm{d}x}\\ &\le\left(\iint|f(x-y)|^p|g(y)|^q\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/r}\\ &\times\left(\iint|g(y)|^q|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/p}\\ &\times\left(\iint|f(x-y)|^p|h(x)|^r\;\mathrm{d}y\;\mathrm{d}x\right)^{1-1/q}\\ &=\left(\int|f(x)|^p\;\mathrm{d}x\right)^{1/p}\left(\int|g(x)|^q\;\mathrm{d}x\right)^{1/q}\left(\int|h(x)|^r\;\mathrm{d}x\right)^{1/r} \end{align} $$