There is a remark following two results in a book that I'm reading that I'm having trouble following.
Let convolution be defined for real- or complex-valued measurable functions on some locally compact topological group $G$ equipped with a left-invariant Haar measure μ. The definition of convolution between two such functions $f$ and $g$ is as follows:
$$f \star g = \int_G f(y)g(y^{-1}\cdot x)\,d\mu(y)$$
Given this definition there are two results:
and
This is followed by a remark I'm having trouble following.
I know that if $g$ in $L^q$ then we can use Theorem 1.5 because $g \in L^q$ implies $g$ is also in weak $L^q$. So it seems to me that Lemma 1.4 implies Theorem 1.5. But they seem to be saying the opposite: that Theorem 1.5 implies Lemma 1.4 up to a constant. Am I missing something?
Under the assumptions of Theorem 1.5, additionally assume that $g\in L^q(G,\mu)$. Then, by Theorem 1.5 and equation (1.3) of Remark 1.6 we infer \begin{align} \Vert f\ast g\Vert_{L^r(G,\mu)} \leq C \Vert f \Vert_{L^p(G,\mu)} \Vert g \Vert_{L^q_w(G,\mu)} \leq C \Vert f \Vert_{L^p(G,\mu)} \Vert g \Vert_{L^q(G,\mu)} \quad(\ast) \end{align} for all $(p,q,r)$ satisfying (1.2), which is the statement of Lemma 1.4 (up to a multiplicative constant). Having only Lemma 1.4 at hand, we know that \begin{align*} \Vert f\ast g\Vert_{L^r(G,\mu)} \leq \Vert f \Vert_{L^p(G,\mu)} \Vert g \Vert_{L^q(G,\mu)}. \quad(\ast\ast) \end{align*} Now, to obtain the statement of Theorem 1.5, we would need to know that $\Vert f \Vert_{L^p(G,\mu)} \Vert g \Vert_{L^q_w(G,\mu)}$ lies in between the left and right hand side of $(\ast\ast)$. But since (1.3) is an upper bound on the weak $L^q$ norm in terms of the $L^q$ norm, that is not something know, and is in general false, since the $L^q$ and the weak $L^q$ norm are not equivalent, as the name also suggests. I hope that clarifies it.