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I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,

Let $1 \le q \le p \le \infty$. Prove that the canonical injection from $L^p(0, 1)$ to $L^q(0, 1)$ is continuous but not compact.

There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it? Is there a simpler approach that does not rely on below Lemma?

Consider $J:L^p(0, 1) \to L^q(0, 1), f \mapsto f$. For $f \in L^p(0, 1)$, we have $\|f\|_q \le \|f\|_p$ and thus $\|J\| \le 1$. Then $J$ is linear continuous. Let's prove that $J$ is not compact. We need the following result, i.e.,

Lemma Let $p \in [1, \infty]$ and $T>0$. Let $f:\mathbb R \to \mathbb R$ be a $T$-periodic function with $$ f(x) := \begin{cases} 0 &\text{if} \quad x \in [0, \frac{T}{2}), \\ 2 &\text{if} \quad x \in [\frac{T}{2}, T). \\ \end{cases} $$ We define $f_n:(0, 1) \to \mathbb R$ by $f_n (x) := f(nx)$ for all $x \in (0, 1)$. Let $\bar f := \frac{1}{T} \int_0^T f (t) \, dt$. We have

  • (1) $f_n \to \bar f$ in in the weak topology $\sigma(L^p, L^{p'})$ where $p'$ is the Hölder conjugate of $p$, and
  • (2) $\lim_n \|u_n-\bar f\|_p = \bigg ( \frac{1}{T} \int_0^{T} |f-\bar f|^p \bigg)^{1/p}$.

Let $f_n$ be given by above Lemma. Then $$ \|f_n\|_q \le \|f_n\|_p \le\|f_n\|_\infty=2 \quad \forall n \ge 1. $$

We claim that $(f_n)$ does not have any convergent sequence in $L^q(0, 1)$. Assume the contrary that $(f_{n_k})_k$ is a subsequence of $(f_n)$ and $g \in L^q(0, 1)$ such that $\|f_{n_k} - g\|_q \xrightarrow{k \to \infty} 0$. Then $f_{n_k} \xrightarrow{k \to \infty} g$ in $\sigma(L^q, L^{q'})$. Then $g= \bar f$ a.e. by (1). On the other hand, $\|f_{n_k} - g\|_q \not \to 0$ as $k \to \infty$ by (2). This is a contradiction. This completes the proof.

Akira
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    If $L^p$ are complex valued the sequence $f_n(x)=e^{2\pi nx}$ is convenient as it does not admit a convergent subsequence in $L^1$ hence in any $L^p.$ Otherwise take $f_n(x)=\sin(2\pi nx).$ The point is that these sequences belong to $L^\infty$ and are orthogonal in $L^2.$ – Ryszard Szwarc Jun 17 '23 at 23:16
  • @RyszardSzwarc Thank you so much for your example. It seems periodic function is the only way to go. – Akira Jun 18 '23 at 06:39
  • There is a slight technical problem in the real case. I will post it as an answer. – Ryszard Szwarc Jun 18 '23 at 06:54

1 Answers1

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Consider $f_n(x)=\sin(2\pi nx).$ As the functions are bounded, it suffices to show that there is no accumulation point in $L^1,$ as $\|f\|_1\le \|f\|_p$ for any $1\le p\le \infty.$ For $n>m$ let $k=n-m$ and $l=n+m.$ Then $$\|f_n-f_m\|_1=\int\limits_0^1|f_n(x)-f_m(x)|\,dx \\ =2\int\limits_0^1|\sin (2\pi kx)|\,|\cos (2\pi lx)|\,dx\\ \ge 2\int\limits_0^1\sin^2(2\pi kx)\,\cos^2 (2\pi lx)\,dx\\ ={1\over 2}\int\limits_0^1[1-\cos (4k\pi x)]\,[1+\cos(4l\pi x)]\,dx={1\over 2} $$

Ryszard Szwarc
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