Let $f\in C_0^\infty(T^d)$, $T$ is a 1-D torus("$0$" here means that integral of $f$ is $0$).
$f=\sum_n a_n\phi_n$, where $\{\phi_n\}$ are $e^{i\vec{n}\cdot\vec{x}}$. $\lambda_n=4\pi |\vec{n}|^2$ are the corresponding eigenvalues of Laplacian. From the regularity of $f$ we know the infinite series must absolutely uniform converge to $f$. Let $\mu_n=\frac{1}{\lambda_n}$.
We consider this operator $A\colon C_0^\infty(T^d)\to C(T^d)$: $$Af = \sum_n \mu_n a_n \phi_n$$ (It's obvious that $Af\in C(T^d)$, since the series absolutely uniform converge)
My question is: does there exist $C>0$, s.t. $$\|Af\|_{L^\infty} \le C \|f\|_{L^\infty}$$
If it doesn't hold, can we modify the right hand side to control the $L^\infty$ norm of $Af$? Thanks!
Let the torus be $T=[0, a)$. The action of operator $A$ is to integrate twice. Indeed, by definition, $$\frac{d^2\phi_n}{dx^2} = \lambda_n \phi_n$$ Thus for $x\in [0, a)$, integrating twice: $$\mu_n\phi_n(x)=\int_0^x\int_0^y \phi_n(u)du\, dy$$ To determine the integration constants, we used the fact that for $n\neq 0$, the average of $\phi_n$ on $T$ is $0$. $$ \begin{split} Af(x)&=\sum_{n\neq 0}a_n\int_0^x\int_0^y \phi_n(u)du\,dy\\ &=\int_0^x\int_0^y \sum_{n\neq 0}a_n\phi_n(u)du\,dy\\ &=\int_0^x\int_0^y f(u)du\,dy \end{split} $$ where you can exchange integrals and sums due to the uniform convergence. From there, you can see that $$\|Af\|_\infty \leq a^2 \|f\|_\infty$$