Let $L^{p,\infty}$ be the weak $L^p$ space consisting of measurable functions $f$ satisfying \begin{equation*} ||f||_{p,\infty}:=\sup_{\rho}\rho\lambda (|f|>\rho)^{\frac{1}{p}}<\infty . \end{equation*}
Given that we have the embedding $L^p(\mathbb{T}^n)\subset L^{p,\infty}(\mathbb{T}^n)$ with $\mathbb{T}^n$ the $n-$dimensional torus, then how come this is equivalent to writing \begin{equation*} ||\cdot||_{p,\infty}\leq ||\cdot||_p? \end{equation*} The inequality suggests that the $L^p$ norm is controlling the behaviour of the weak $L^p$ norm but does the embedding say that weak $L^p$ is controlling the behaviour of $L^p$? Why is the inequality not the other way around?
This is just an example, but I have seen this with many other inequalities and embeddings.
This may be trivial, but I just wanted some clarification. Thank you in advance.
To address your second and third questions, note that the inequality $\| \cdot\|_{p,\infty} \le \|\cdot\|_p$ leads to $$ \|\cdot\|_p < \infty \implies \|\cdot\|_{p,\infty} < \infty$$ which in turn gives $$f \in L^p \implies f \in L^{p,\infty}$$ i.e., $L^p \subset L^{p,\infty}$. That is why the inequality seems backward from the containment.
How does the inequality arise from the containment? If $X \subset Y$ are Banach spaces and the identity $I : X \to Y$ is continuous, then there exists $C$ satisfying $\|f\|_Y \le C \|f\|_X$.