Assume $E$ has finite measure and $1\leq p < \infty$. Suppose $f_{n}$ is a sequence of measurable function that converges pointwise a.e on $E$ to $f$. Show that $f_{n} \to f$ in $L^{p}(E)$, if there is a $\theta> 0$ such that $f_{n}$ belongs to and is bounded as a subset of $L^{p+\theta}(E)$.
I saw a solution in the internet which goes as follows :
Suppose that there exists $\theta>o$ such that $||f_{n}||_{p+\theta} < C$ for all $n$. By Fatou's Lemma we have $f\in L^{p+\theta}(E)$ and $||f||_{p+\theta} < C$.
Now, we have by an application of Holder's inequality:
$$\int_{E}|f-f_{n}|^{p} \leq m(E)^{\frac{\theta}{p+\theta}}\int_{E}|f_{n}-f|^{p+\theta} \leq 2C^{p+\theta} m(E)^{\frac{\theta}{p+\theta}}$$ So, $\{\ |f_{n}-f|^{p} \}\ $ is uniformly integrable and hence by Vitali convergence theorem $f_{n}\to f$ in $L^{p}(E)$. What I want to know is that how is the Holder's inequality being used here?
Thanks in advance!
Consider an arbitrary $F\subset E$. Let $u = 1_F$ and $v = \lvert f-f_n\rvert^p$. Then, $$\|uv\|_1\leq \|u\|_{1+p/\theta}\|v\|_{1+\theta/p}$$ by Hölder's inequality. Here, $$\|u\|_{1+p/\theta} = \left[\int_F 1\,\mathrm{d}x\right]^{\frac{\theta}{p+\theta}} = m(F)^{\frac{\theta}{p+\theta}}$$ and $$\|v\|_{1+\theta/p} = \left[\int_E \lvert f-f_n\rvert^{p(1+\theta/p)}\,\mathrm{d}x\right]^{\frac{p}{p+\theta}} = \left[\int_E \lvert f-f_n\rvert^{p+\theta}\,\mathrm{d}x\right]^{\frac{p}{p+\theta}} = \|f-f_n\|_{p+\theta}^p$$ Given that we have some upper bound for $\|f-f_n\|_{p+\theta}^p$, for any $\epsilon > 0$ we can choose some $\delta > 0$ such that $m(F) < \delta$ guarantees that $\int_F \lvert f-f_n\rvert^p\,\mathrm{d}x < \epsilon$, which is the uniform integrability condition for $\{\lvert f-f_n\rvert^p\}$.