Let $K$ be compact set on $\mathbb{R}^n$ with non-empty interior. Assume that $p \geq 2 $ and that it is given a sequence of measurable functions $\{u_n : K \to \mathbb{R}\} \subset L^{p-\epsilon}(K)\cap L^p(K)$, for each $\epsilon > 0$, such that there is $u\in L^p(K)$ such that $$\lim_{n\to\infty}\int_K|u_n-u|^{p-\epsilon} = 0.$$
My question is: is it true that $$\lim_{n\to\infty}\int_K|u_n-u|^p = 0?$$
I believe so. These are my thoughts: take $\left\{\dfrac{1}{n} : n > 0\right\}$ and consider the function $$\Phi : \bigcup_{n\geq 1}L^{p-\frac{1}{n}}(K)\cap L^ p(K)\times \bigcup_{n\geq 1}[p-\frac{1}{n},p]\to \mathbb{R}$$ $$(f,q) \mapsto \|f\|_q.$$
I think that since $\mu(K) < \infty$ then $\Phi$ is continuous, so the result should follows.
Is this correct?
This is false. Consider $K = [0,1]$ and let $u_n = n 1_{[0,n^{-p})}$. Then $$\int_K |u_n|^{p-\varepsilon} = n^{-\varepsilon} \to 0$$ for any $\varepsilon > 0$ but $$\int_K |u_n|^p = 1.$$