$L^{p}$ spaces and their properties

Question

I have a question: I don't know how to show

that if $1<p<q<\infty$ , then $L^{q}(0,1)\subset L^{p}(0,1)$ and $\mid\mid f\mid\mid_p$ < $\mid\mid f\mid\mid_q$, $\,f \in L^{q}(0,1)$?

Answer 1

Hint: apply the Holder inequality with the parameters $\frac qp > 1$ and $$ \frac{\frac qp}{1-\frac qp} $$

Answer 2

If $1<p<q<\infty$, and $f\in L^q(0,1)$, then $$ \int_0^1 \lvert\, f(x)\rvert^p\,dx=\int_0^1 \lvert\, f(x)\rvert^p\cdot 1\,dx \le \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}\left(\int_0^1 1^r\,dx\right)^r= \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}, $$ where $$ \frac{p}{q}+\frac{1}{r}=1\quad\text{or}\quad r=\frac{q}{q-p}. $$ Hence $$ \left(\int_0^1 \lvert\, f(x)\rvert^p\,dx\right)^{1/p}=\|\,f\|_p\le \|\,f\|_q = \left(\int_0^1 \lvert\, f(x)\rvert^q\,dx\right)^{1/q}. $$ Therefore, if $f\in L^q(0,1)$, then $\int_0^1\lvert\,f(x)\rvert^p\le \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}<\infty$, and hence $f\in L^p(0,1)$. Therefore, $$ L^q(0,1)\subset L^p(0,1). $$