Let $G$ be an algebraic group over a field $F$, and let $(\pi,V)$ be a smooth $G$-representation over an algebraic closed field $k$. Then $\pi$ is called a CUSPIDAL representation if $r(V)=0$ for any Jacquet functor $r$.
Equivalently, $\pi$ is CUSPIDAL if it is not isomorphic to an irreducible quotient of any parabolically induced representation of $G$.
On the other hand, $\pi$ is SUPCUSPIDAL if it is not isomorphic to an irreducible subquotient of any parabolically induced representation of $G$.
I find that some experts claim the following thing.
Suppose that the characteristic of $F$ is $0$. Moreover, if $F$ is a $p$-adic field, then $k\neq\bar{\mathbb{Q}}_p$. Then CUSPIDAL is equivalent to SUPCUSPIDAL!
I feel surprised. How can subquotient automatically be quotient under this assumption?
This is a fairly non-trivial fact, and I only know know to prove it using (parts of -- namely the splitting of the supercuspidal part of the category) the Bernstein decomposition. Let $(M,\sigma)$ be a supercuspidal support, and let $\pi$ be an irreducible subquotient of $i_M^G \sigma$. Then $r_M^G \pi$ is a finite length admissible representation containing $\sigma$ as a subquotient. Write $\tau=r_M^G \pi$, and split $\tau=\tau_{nc}\oplus\tau_c$ into non-cuspidal and cuspidal parts, then further split $\tau_c=\bigoplus_{\sigma_i}\tau_\sigma$ according to the cuspidal inertia classes. If this is unfamiliar to you, let $^0M$ denote the subgroup of $M$ generated by all of its compact open subgroups, and then $\tau_{\sigma_i}$ is the maximal $\sigma_i|_{^0M}$-isotypic subrepresentation of $\tau_c$.
Since $\sigma$ is a subquotient of $\tau$, it must be a subquotient of $\tau_\sigma$ (it has no non-zero maps into or from $\tau_{\sigma'}$ for $\sigma'$ defining an inertia class other than that of $\sigma$). Now, by definition, any irreducible subquotient of $\tau_\sigma$ is of the form $\sigma\otimes\omega$ for some unramified character $\omega$ of $M$ (this is precisely what restriction to $^0M$ does). Our claim is that $0\neq Hom_M(\tau_\sigma,\sigma)$.
Let $K$ be a compact open subgroup of $M$ such that $\sigma$ has a non-zero vector fixed by $K$. $K$ is compact, so unramified twists do nothing on $K$, hence $\tau_\sigma$ also has a non-zero vector fixed by $K$. The representation $\tau_\sigma$ is of finite length, since it's a summand of the finite length representation $\tau$. Restricting to $^0M$, $\tau_\sigma$ and $\sigma$ are both compact representations, hence semisimple, and so we have $0<\dim Hom_{^0M}(\tau_\sigma,\sigma)<\infty$. The group $G/{}^0G$ acts on this space in the usual way, and the space $Hom_M(\tau_\sigma,\sigma)$ is the space of fixed points of this action.
Since $\sigma$ is a subquotient of $\tau_\sigma$, there exist subrepresentations $\tau_1,\tau_2$ of $\tau_\sigma$ such that $\tau_2/\tau_1\simeq\sigma$. $\sigma$ is a finite length compact representation of $^0M$, hence injective as an $^0M$-module, so there is an exact sequence $$0\rightarrow Hom_{^0M}(\tau_1,\sigma)\rightarrow Hom_{^0M}(\tau_2,\sigma)\rightarrow Hom_{^0M}(\tau_2/\tau_1,\sigma)\rightarrow 0.$$ This sequence carries a term-wise action of $M/{}^0M$, and so this is an exact sequence of $M/{}^0M$-modules. From the inclusion $\tau_2\hookrightarrow\tau$, we get an embedding $Hom_{^0M}(\tau_2,\sigma)\rightarrow Hom_{^0M}(\tau_\sigma,\sigma)$.
So $Hom_{^0M}(\tau_2/\tau_1,\sigma)$ is a subquotient of $Hom_{^0M}(\tau_\sigma,\sigma)$ with a non-zero $M/{}^0M$-fixed point. Then I claim (and will leave to you to check) that since $Hom_{^0M}(\tau_\sigma,\sigma)$ is a finite-dimensional $M/{}^0M$-module admitting a subquotient $Hom_{^0M}(\tau_2/\tau_1,\sigma)$ with a non-zero fixed point, the module $Hom_{^0M}(\tau_\sigma,\sigma)$ has a fixed point, which is to say that $Hom_M(\tau_\sigma,\sigma)\neq 0$.
The same argument works the other way, and you see that $hHom_M(\tau_\sigma,\sigma)\neq 0$ if and only if $Hom_M(\sigma,\tau_\sigma)\neq 0$. Then, by adjunction, you get what you wanted.
I couldn't easily tell you where this goes wrong in positive characteristic, since it builds on a whole load of big theorems, but the standard example of a representation which is cuspidal but not supercuspidal is to look at $\bar{\mathbb{F}}_3$-representations of $GL_3(\mathbb{Q}_2)$, and consider the representation parabolically induced from the trivial representation of the split maximal torus. This is indecomposable of length 3, and it has a unique subrepresentation and quotient, which are the trivial character and the unramified quadratic character respectively. The subquotient in the middle is then cuspidal, but not supercuspidal.