Canonical injection from $L^p(0, 1)$ into $L^q(0, 1)$ continuous? How about compact?

Question

Let $1 \le q \le p \le \infty$. I have two questions.

1. Is the canonical injection from $L^p(0, 1)$ into $L^q(0, 1)$ continuous?
2. Is the canonical injection from $L^p(0, 1)$ into $L^q(0, 1)$ compact?

Answer 1

First problem: By Jensen,

$$ (\int_0^1|f|^q\,)^{p/q} \le \int_0^1 |f|^p.$$

Taking $p$th roots gives $\|f\|_q \le \|f\|_p.$ The inclusion map from $L^p$ to $L^q$ is thus bounded and linear, hence is continuous.

Second problem: Consider the functions $e^{inx}.$ These functions lie in the unit sphere of $L^p.$ If the inclusion map were compact, then some subsequence of $e^{inx}$ would converge in $L^q.$ Hence some further subsequence would converge a.e. That this can't happen follows from

Thm: Let $n_1 < n_2 < \cdots $ be a sequence of integers. Then $e^{in_kx}$ diverges for a.e. $x\in \mathbb R.$

Proof: Let $E = \{x\in [0,2\pi]: e^{in_kx}\text { converges }\}.$ It's enough to show $m(E)=0.$ For $x\in E,$ let $f(x) = \lim e^{in_kx}.$ Then $|f|\equiv 1$ on $E.$ By the dominated convergence theorem,

$$m(E) = \int_E 1 = \int_E |f|^2 = \int_E \bar f f= \int_E \overline {f(x)}\cdot (\lim e^{in_kx})\, dx = \lim \int_E \overline {f(x)} e^{in_kx}\, dx. $$

By the Riemann Lebesgue lemma, the last limit is $0$ and we're done.

Answer 2

Here are the answers:

1. Yes.

2. No.