Consider function $L^p(X)$ and $L^q(X)$ space with $p<q$ where $X$ is a finite measure space. Then from Holder, I have $L^q(X)$ embedding into $L^p(X)$. In particular, if $\|f\|_q<\infty$, then $\|f\|_p<\infty$. In particular, $L_q\subset L_p$
Now consider $BV_p([0,1],\Bbb R)$ with where $BV_p([0,1],\Bbb R)$ is space of functions from $[0,1]$ to $\Bbb R$ s.t. p-th variation of any function is finite. Note p-th variation requires taking over sup over partition and denote p-th variation as $v_p(f)$. Then one could show for $1<p<q$, if $v_p(f)<\infty$, then $v_q(f)<\infty$. In particular, $BV_q([0,1],\Bbb R)\subset BV_p([0,1],\Bbb R)$
Q: It seems that in bounded variation case, the embedding is in the same direction. What leads to this same direction inclusion? I am looking for intuitive explanation for inclusion. The proof in bounded variation case does not use Holder inequality. However, p-th variation looks very much like p-norm in $L_p$ space. That is why I would hope the inclusion follows the same direction.
This comes from the fact that the $v_p$ norm is defined via $\ell_p$-norms, that is, norms on finite dimensional vectors of the form $\left(f\left(t_i\right)-f\left(t_{i-1}\right)\right)_{i=1}^n$, where $0=t_0<t_1<\dots<t_n\leqslant 1$. Therefore, one has $$ \left(\sum_{i=1}^n \left\lvert f\left(t_i\right)-f\left(t_{i-1}\right)\right\vert^p \right)^{1/p}\leqslant \left(\sum_{i=1}^n \left\lvert f\left(t_i\right)-f\left(t_{i-1}\right)\right\vert^q \right)^{1/q} $$ and since $$ \left(\sum_{i=1}^n \left\lvert f\left(t_i\right)-f\left(t_{i-1}\right)\right\vert^q \right)^{1/q}\leqslant v_q(f), $$ one has for each partition $0=t_0<t_1<\dots<t_n\leqslant 1$ that $$ \left(\sum_{i=1}^n \left\lvert f\left(t_i\right)-f\left(t_{i-1}\right)\right\vert^p \right)^{1/p}\leqslant v_q(f) $$ hence $v_p(f)\leqslant v_q(f)$.