$L^p$ spaces and proper inclusion

Question

Let $1≤p < q$. Prove that $L^p(\mathbb{R}) \subset L^q(\mathbb{R})$ and the inclusion is proper.

I am unsure how to begin this or even prove it about $L^p$ spaces and Banach spaces.

Answer 1

Let $f(x):= x^{-1/p} \mathbf{1}_{(1,\infty)}$. Then $f^q$ is integrable because $\frac{q}{p} > 1$. But $f^p = \frac{1}{x}$ (on $(1,\infty)$) is not.

That is, $f \in L^q(\mathbb{R})$ but $f \notin L^p(\mathbb{R}) $.

Answer 2

As Frank provided a counterexample the inclusion in general doesn't hold. In fact if your measure is finite the inclusion is vice versa because via Hölder you get \[ \|f\|_q \leq \mu(X)^{\frac{1}{p} -\frac{1}{q}} \cdot \|f\|_p.\] Maybe you are talking about the sequence spaces $\ell^p$ because there the inclusion holds as every function which is in $\ell^p$ converges to zero, so it will be smaller than $1$ in some cofinite set. But as $p\mapsto x^p$ is monotone decreasing for $x\in [0,1]$ you can dominate the integral (i.e. the series).