Let $\Omega$ be open bounded set, $1 < s < q < \infty$, $C>0$, Let $\{u_k\} \subset L^q(\Omega)$ with a following properties:
$$ ||u_k||_s \to 0, \quad ||u_k||_q < C$$
Does it hold that $u_k \to 0$ in $L^q$ ($||u_k||_q \to 0$)? Is a $||u_k||_q < C$ necessary to show that?
No, this does not hold. For example, consider the function $$f_n(x)=\left\{\begin{array}{c l} n\sqrt{n}x, & x\in\left[0,\frac{1}{n}\right] \\ 2\sqrt{n}-n\sqrt{n}x, & x\in\left(\frac{1}{n},\frac{2}{n}\right]\end{array}\right.. $$ Then, we compute $$\int_0^1|f_n(x)|^{3/2}\,dx=\int_0^{1/n}(n\sqrt{n}x)^{3/2}\,dx+\int_{1/n}^{2/n}\left(2\sqrt{n}-n\sqrt{n}x\right)^{3/2}\,dx,$$ and if we set $y=2/n-x$ in the second integral, we have that $$\begin{align*}\int_0^1|f_n(x)|^{3/2}\,dx&=\int_0^{1/n}(n\sqrt{n}x)^{3/2}\,dx+\int_0^{1/n}(n\sqrt{n}y)^{3/2}\,dy\\ &=2(n\sqrt{n})^{3/2}\cdot\frac{2}{5}n^{-5/2},\end{align*}$$ which goes to $0$ as $n\to\infty$. Hence, $\|f_n\|_{L^{3/2}(0,1)}\to 0$. However, the sequence $\left(\|f_n\|_2\right)_n$ is bounded and does not go to $0$.