= Solution
Use normalized <Haar measure> $dm=dt/(2\pi)$. A <homogeneous Banach space on the circle> is a Banach subspace $B\subseteq L^1(\mathbb T)$ with $\|f\|_1\leq\|f\|_B$, such that translations $\tau_s f(t)=f(t-s)$ preserve $B$, satisfy $\|\tau_sf\|_B=\|f\|_B$, and obey $\|\tau_sf-f\|_B\to0$ as $s\to0$.
For continuous $g$, the map $s\mapsto g(s)\tau_sf$ is continuous with values in the complete space $B$. Define its Banach-valued Riemann integral, equivalently its <Bochner integral>, by
$$
f*g=\int_{\mathbb T}g(s)\tau_sf\,dm(s).
$$
The <triangle inequality> for Riemann sums and passage to their norm limit give the canonical estimate
$$
\boxed{\|f*g\|_B\leq\int_{\mathbb T}|g(s)|\|\tau_sf\|_B\,dm(s)=\|f\|_B\|g\|_1.}
$$
The continuous inclusion into $L^1$ carries the same Riemann sums to the ordinary <convolution>, so this definition agrees <almost everywhere> with $\int f(t-s)g(s)\,dm(s)$. <Density> of <continuous functions> in $L^1$, <completeness> of $B$, and the estimate extend the operation uniquely to all $g\in L^1$. The ordinary $L^1$ <convolution> estimate identifies that extension with the same integral formula.
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