Let $p, h, d\in\mathbb{R}, h>0, p>1$ and $n\in\mathbb{N}, n\ge 2$. Let $(x_n)_n\subset (W_0^{1, p}(\Omega),\|\cdot\|)$. During the class of today, the lecturer said that the inequalities $$\|x_n\|^p\left(h^{\frac{n}{n-p}}-\|x_n\|^{\frac{p^2}{n-p}}\right)\le o(1)$$ and $$\|x_n\|^p\le n(d +c_1)+o(1)$$ for a real positive constant $c_1$, ensure that $x_n\to 0$ in $W_0^{1, p}(\Omega)$ provided that $$d<\frac{h^{\frac{n}{p}}}{n}-c_1.$$ Actually, I don't understand why. Could someone please help me to justify that?
Thank you in advance.
Let $$d+\epsilon=\frac{h^\frac{n}{p}}{n}-c_1$$ By construction, $\epsilon>0$.
Substituting into your second bound, we have $$\|x_n\|^p\leq n(d+c_1)+o(1)=h^{\frac{n}{p}}-\epsilon n+o(1)$$ Raising to $\frac{p}{n-p}$th power, \begin{align*} \|x_n\|^{\frac{p^2}{n-p}}&\leq(h^{\frac{n}{p}}-\epsilon n+o(1))^{\frac{p}{n-p}} \\ &=h^{\frac{n}{n-p}}\left(1-\frac{p}{n-p}\cdot\epsilon n+o(1)\right) \\ &=h^{\frac{n}{n-p}}\left(1-\frac{\epsilon}{\frac{1}{p}-\frac{1}{n}}\cdot(1+o(1))\right) \end{align*} Thus $$h^{\frac{n}{n-p}}-\|x_n\|^{\frac{p^2}{n-p}}\geq h^{\frac{n}{n-p}}\cdot\frac{\epsilon}{\frac{1}{p}-\frac{1}{n}}\cdot(1+o(1))=\Omega(1)$$ (using Knuth's "Big Omega").
Finally, rearranging your first bound then gives $$\|x\|^p\leq\frac{o(1)}{\Omega(1)}$$ as required.