Let $μ(x)<∞$ and $1\leq p\leq q<∞$. Show that $L^q \subset L^p$ and $f_n \to f$ in $L^p$ implicates $f_n \to f$ in $L^q$.
Solution: Let $f \in L^q$.If $μ(x)<∞$, then the constant function with value $1$ is in $L^s$ for every $1\le s$ . From Hölder's inequality for $\frac{q}{p}>1$ we have
$$\int |f(x)|^p* 1\,dμ(x)\leq \left(\int |f(x)|^p\,dμ(x)\right)^\frac{p}{q} * \left(\int 1\,dμ(x)\right)^\frac{q-p}{q}$$
That is $||f||_p \leq ||f||_q *μ(x) ^\frac{q-p}{qp} <∞$
Therefore $f \in L^p$.
Question: How do I show the second part of the exercise that is the implication? Doesn't it apply that if $L^q \subset L^p$ and $f_n \to f$ in $L^p$ then $f_n \to f$ in $L^q$?
It's not true. There is not even a way to guarantee that $(f_n)$ and $f$ are elements of $L^q$.
For instance, consider $p=1$ and $q=2$, and let $f$ be defined on $[0,1]$ by $f(x)=\sqrt{x}$. Then taking $f_n=f$ for each $n$, we have $f_n\to f$ in $L^1$ but $f_n,f\not\in L^2$ for every $n$.
The correct implication is $f_n\to f$ in $L^q$ implies $f_n\to f$ in $L^p$. The proof of this is easy using what you have already shown: If $f_n\to f$ in $L^q$, then $\|f_n-f\|_q\to 0$ so that $$ \|f_n-f\|_p\le\|f_n-f\|_q\mu(X)^{\frac{q-p}{qp}}\to 0. $$