$\int _{\mathbb{T}^n}f(t)g(x-t)d\lambda (t)\in\mathbb{C}$ for all $x\in\mathbb{T}^n$ and $f,g\in L^1(\mathbb{T}^n)$?

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According to the book "Classical Fourier Analysis" (written by Loukas Grafakos) every locally compact group has a Haar measure. Consider the set $\mathbb{T}^n:=\mathbb{R}^n/(2\pi\mathbb{Z}^n)=\big\{x+2\pi\mathbb{Z}^n:x\in\mathbb{R}^n\big\}$ with the metric defined by $d(x+2\pi\mathbb{Z}^n,y+2\pi\mathbb{Z}^n):=\inf \big\{\Vert x-y+2\pi j\Vert _n:j\in\mathbb{Z}^n\big\}$. Then $\mathbb{T}^n$ is a locally compact group with respect to that metric. Therefore $\mathbb{T}^n$ has a Haar measure.

Suppose that $\lambda$ is the Haar measure of $\mathbb{T}^n$ such that $\lambda (\mathbb{T}^n)=1$.

My question is: $\int _{\mathbb{T}^n}f(t)g(x-t)d\lambda (t)\in\mathbb{C}$ for all $x\in\mathbb{T}^n$ and $f,g\in L^1(\mathbb{T}^n)$? That is, is the convolution product of $f$ and $g$ well-defined for all $f,g\in L^1(\mathbb{T}^n)$?

Unfortunately I have made no progress in trying to answer the above question.

Thank you for your attention!

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Let $f$ be any function in $L^{1}$ which is not in $L^{2}$. Take $g(y)=f(-y)$ so that $g$ is also in $L^{1}$. Then then convolution does not exist at $x=0$.