Compactness of $L^p$ inclusion into $L^q$

Question

Let $ i \colon L^p (0, 1) \longrightarrow L^q (0, 1) $, when $ p \ge q $ the canonical inclusion. It is clearly continuos but never compact: I cannot succeed in showing the last point. Thanks for the answers

Answer 1

It is not compact by the Khintchine inequality which tells you that the closed span of Rademacher functions is isomorphic to a Hilbert space in every $L_p$-space. The inclusion then takes a copy of a Hilbert space to another copy of a Hilbert, so it cannot be compact.

Answer 2

You can take $\{\sin(nx)\}_{n\geq 1}$. It is bounded in $L^p$ but you can't take a sub-sequence converging in $L^q$ because $\sin(nx)$ has no sub-sub sequences converging a.e. (Remind that strong convergence in $L^q$ implies a. e. convergence up to taking subsequences).

Answer 3

Argue by contradiction: by composing with the two inclusions $L^\infty\to L^p$ and $L^q\to L^1$ you would get a compact inclusion $L^\infty\to L^1$.
But now let $f_0(x)=1$ if $\lfloor x\rfloor$ (the integer part) is even and $f_0(x)=0$ otherwise, for any $x\in\mathbb{R}$. Let $f_n(x)=f_0(2^n x)$.
The restriction to $(0,1)$ of the functions $f_n$ forms a bounded sequence in $L^\infty(0,1)$ with no converging subsequence in $L^1(0,1)$ (in fact if $m\neq n$ you have $\|f_n-f_m\|_1=\frac{1}{2}$), which contradicts the compactness of $L^\infty\to L^1$.