A function $f(x)$ is called low-dimensional if there exists non-zero vector $v\in R^n$ such that $f(x)=f(x+cv)$ for all $c\in R$. I'm wondering whether any finite sum of continuous low-dimensional functions $f\_i(x)$ is either non-integrable or being the zero function. It's trivially true for $n=1$ and if we restrict $f\_i(x)$ to be piece-wise affine it also holds for $n=2$ by considering non-differentiable regions.


I have no clue how to deal with $n>2$ cases, even if we restrict $f\_i(x)$ to be piece-wise affine. The Radon transform seems tempting however each $f\_i(x)$ might be non-integrable even if we assume by contradiction $f(x)$ is integrable. What mathematical tools may help solving this problem?



Finite sum of low-dimensional functions in R^n
I assume that by integrable you mean $\int |f|<\infty$.


Assume that $f=f\_1+\ldots+f\_n$ where each $f\_i$ is a continuous periodic function: $f\_i(x+v\_i)=f\_i(x)$ for certain $v\_i\in \mathbb{R}^n\setminus \{0\}$. This is weaker condition then being low-dimensional. I claim that if $\int |f|<\infty$ then $f\equiv 0$.


Induction in $n$. Base $n=1$: denote $U\_i=H+[i,i+1)\cdot v\_1$, $i\in \mathbb{Z}$, where $H$ is the hyperplane orthogonal to $v\_1$. The integrals over $U\_i$ of $|f|=|f\_1|$ are equal by periodicity, thus they all must be equal to 0 and $f=f\_1\equiv 0$. Induction step: if $f$ is integrable, so is $g:=f(x+v\_n)-f(x)=\sum\_{i=1}^{n-1} g\_i(x)$, $g\_i(x):=f\_i(x+v\_n)-f\_i(x)$. Each $g\_i$ is $v\_i$-periodic, thus by induction we have $g\equiv 0$. Hence $f$ is $v\_n$-periodic and we are in the base case situation.


