Let $\varphi \in C^{\infty}(\mathbb{T}^n,\mathbb{C})$ (i.e. just smooth periodic complex-valued function) and $f \in C^{\infty}(\mathbb{T}^n,\mathbb{C}^m)$. Then I was wondering if the following is true:
$$ \|\varphi f\|_{2} \leq \|\varphi\|_{\infty}\|f\|_{2} $$. Where the $\|\cdot \|_{2}$ is $L^2-$norm and $\|\cdot\|_{\infty}$ is the sup-norm which are defined in this case to be
$$ \|f\|_{2} := \frac{1}{(2\pi)^{\frac{n}{2}}} \Bigg(\int_{\mathbb{T}^n} f \cdot f \Bigg)^{1/2} \quad \text{where $'\cdot'$ represents the Hermitian product.} $$ and $$ \|f\|_{\infty} = \sup_{\mathbb{T}^n}\{|f|\} $$ also we denote $|f| = (f \cdot f)^{1/2}$.
I managed to prove the following $$ \|f\|_2 \leq \|f\|_{\infty} $$ as follows $$ |f| \leq \sup_{\mathbb{T}^n} |f| $$ $$ \implies |f| \leq \|f\|_{\infty} \implies |f|^2 \leq \|f\|_{\infty}^2 $$ hence $$ \int_{\mathbb{T}^n} |f|^2 d\mu \leq \int_{\mathbb{T}^n} \|f\|_{\infty}^2 d\mu = \|f\|_{\infty}^2 \int_{\mathbb{T}^n} 1 d\mu = \|f\|_{\infty}^2 \mu(\mathbb{T}^n) $$ and taking square roots on both sides, $$ \Bigg(\int_{\mathbb{T}^n} |f|^2 d\mu \Bigg)^{1/2} \leq \|f\|_{\infty} \mu(\mathbb{T}^n)^{1/2} $$ which gives $$ \Bigg(\int_{\mathbb{T}^n} f \cdot f \quad d\mu \Bigg)^{1/2} \leq \|f\|_{\infty} \mu(\mathbb{T}^n)^{1/2} $$ multiplying $\frac{1}{(2\pi)^{n/2}}$ both sides, we obtain $$ \frac{1}{(2\pi)^{n/2}}\Bigg(\int_{\mathbb{T}^n} f \cdot f \quad d\mu \Bigg)^{1/2} \leq \frac{1}{(2\pi)^{n/2}}\|f\|_{\infty} \mu(\mathbb{T}^n)^{1/2} $$ and hence by using the fact that $\mu(\mathbb{T}^n) = (2\pi)^n$, we have $$ \|f\|_{2} \leq \|f\|_{\infty} $$ But this seems not so useful. I was proceeding by showing $$ \|\varphi f\|_2 \leq \|\varphi\|_{2}\|f\|_2 $$ which I think is False even with the given conditions and hence the above work seems useless.
Summary : I need to prove $$ \|\varphi f\|_{2} \leq \|\varphi\|_{\infty}\|f\|_{2} $$.
Remember that integrals are monotonic, so if $f\leq g$ a.e. then $\int f\leq \int g$. Thus if $g\geq 0$ a.e. and $f_1\leq f_2$ a.e. you get $f_1g\leq f_2g$ a.e. and thus $\int f_1g\leq \int f_2 g$.
So now see that $|\phi|\leq \|\phi\|_\infty$ a.e. and $(\phi f)\cdot(\phi f) = |\phi|^2 |f|^2$ and $|f|^2\geq 0$.
So using this we get $$ \|\phi f\|_2 \leq \Big\|\|\phi\|_\infty f\Big\|_2$$ and as $\|\phi\|_\infty$ is a scalar this is $$ \|\phi\|_\infty \|f\|_2$$