Inclusion of Lp into L1 on bounded domains using Holder inequal On bounded domains, $L^p \subset L^1, p>1$ because on such domains a function isn't integrable if the function is too large. Exponentiating the function makes it larger. Is it possible to show that $L^1(a,b)$ space is a more general space than $L^p(a,b),p>1$ using Holder inequality (below)? $$\int_{a}^{b}|f(x)g(x)|dx \leq \Big(\int_{a}^{b}|f(x)|^pdx\Big)^{1/p} \Big(\int_{a}^{b}|g(x)|^qdx\Big)^{1/q}$$

If $f\in L^p(a,b)$ with $p>1$, then using Holder's inequality with $f$ and $1$ we get $$ \int_a^b|f(x)|\;dx\leq\Big(\int_a^b|f(x)|^p\;dx\Big)^{\frac{1}{p}}\Big(\int_a^b\;dx\Big)^{\frac{1}{q}}=(b-a)^{\frac{1}{q}}||f||_p $$ Therefore $f\in L^1(a,b)$, and in fact $$ ||f||_1\leq (b-a)^{\frac{1}{q}}||f||_p$$