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Let $X_0$,$X_1$ and $X_2$ be Banach spaces such that

$$X_2\subset\subset X_1\subset X_0$$

I want to prove that for every $\epsilon>0$, there exists a constant $C_\epsilon>0$ such that for all $x\in X_2$, the inequality

$$\lVert x\rVert_{X_1}\leq\epsilon\lVert x\rVert_{X_2}+C_\epsilon \lVert x\rVert_{X_0}$$ holds
My attempt is to argue by contradiction i.e. suppose that there is $\epsilon>0$ such that for all $C>0$, there is an $x_C\in X_2$ such that $$\lVert x\rVert_{X_1}>\epsilon\lVert x\rVert_{X_2}+C \lVert x\rVert_{X_0}$$ then let $C=k\in\mathbb{N}$, it gives a sequence $\{x_k\}\subset X_2$ satisfying this inequality with $C=k$.
divided both side $\frac{1}{\lVert x\rVert_{X_1}}$ then we can find a bounded sequence in $X_2$ that is $$y_k=\frac{x_k}{\lVert x\rVert_{X_1}}$$ and $y_k$ satisfies $$1>\epsilon\lVert y_k\rVert_{X_2}+k \lVert y_k\rVert_{X_0}$$ so it should have a subsequence which is convergent both in $X_1$ and $X_0$, denoted by $\{z_k\}$
since $X_1$ is Banach so it has a limit point say $z\in X_1$

then I don't know how to continue.
My teacher explains the notions he used $$X_2\subset\subset X_1\subset X_0$$ here means the embedding of $X_2$ into $X_1$ is compact and the embedding of $X_0$ is continuous. In particular, every bounded sequence in $X_2$ has a subsequence which is convergent in $X_1$ and any convergent sequence in $X_1$ is also convergent in $X_0$.

Emiya
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  • What does $\subset \subset$ mean? – Kavi Rama Murthy Nov 03 '21 at 11:29
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    If $\overline{X_2}$ is compact then $X_2={0}$. – David C. Ullrich Nov 03 '21 at 12:58
  • @DavidC.Ullrich, I still have a question. I know that a bounded sequence in $X_2$ should have a subsequence which is convergent in $X_1$ and so convergent in $X_0$. Denote the subsequence by ${y_k}$ and $y$ be its limit in $X_1$. Is ${y_k}$ convergent in $X_2$ and $y$ also in $X_2$? – Emiya Nov 03 '21 at 14:41
  • The only sequence in $X_2$ is $(0,0,0,\dots)$. So $y=0$. – David C. Ullrich Nov 03 '21 at 15:34
  • @DavidC.Ullrich, but why there is only ${0}$ in $X_2$? – Emiya Nov 03 '21 at 18:27
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    @David: I guess what is actually meant is that $X_2$ compactly embeds into $X_1$ (i.e., the unit ball of $X_2$ is relatively compact in $X_1$). – PhoemueX Nov 03 '21 at 20:20
  • @PhoemueX Or in other words, that the inclusion is a compact operator. Yes, that makes more sense. And come to think of it that could well be what was meant by "$\overline {X_2}\subset X_1$ and it is compact"; we merely need to supply a charitable referent for "it". Thanks. – David C. Ullrich Nov 03 '21 at 22:19
  • @Emiya It's possible I've misread the whole thing. I read "$\overline{X_2}\subset X_1$ and it is compact" as saying that $\overline{X_2}$ is compact. The only compact Banach space is ${0}$. – David C. Ullrich Nov 03 '21 at 22:22

1 Answers1

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I think you are almost done so I will just follow your steps.
assume contrary which means $\exists\epsilon>0$ $\forall C>0$ $\exists x_C\in X_2$ such that $$\lVert x_C\rVert_{X_1}>\epsilon\lVert x_C\rVert_{X_2}+C\lVert x_C\rVert_{X_0}$$ let $C=k\in\mathbb{N}$ and $x_k$ be the correspnding $x_C$
then $\{x_k\}$ is a sequence in $X_2$ satisfying $$\lVert x_k\rVert_{X_1}>\epsilon\lVert x_k\rVert_{X_2}+k\lVert x_k\rVert_{X_0}$$ then for each $x_k$ we have $$\lVert x_k\rVert_{X_1}>\epsilon\lVert x_k\rVert_{X_2}+k\lVert x_k\rVert_{X_0}\geq 0+0=0$$ so $\lVert x_k\rVert_{X_1}\neq0$
hence we can define $y_k=\frac{x_k}{\lVert x_k\rVert_{X_1}}$
then $$\frac{1}{\lVert x_k\rVert_{X_1}}\lVert x_k\rVert_{X_1}>\epsilon\frac{1}{\lVert x_k\rVert_{X_1}}\lVert x_k\rVert_{X_2}+k\frac{1}{\lVert x_k\rVert_{X_1}}\lVert x_k\rVert_{X_0}$$ which implies $$1>\epsilon\lVert y_k\rVert_{X_2}+k\lVert y_k\rVert_{X_0}$$ so $\{y_k\}$ is a bounded sequence in $X_2$
so it has a subsequence which is convergent in $X_1$
write $\{z_k\}$ as the convergent subsequence
and since $X_1$ is Banach, this sequence has a limit point say $z\in X_1$
then $\{z_k\}$ should also convergent to $z$ in $X_0$
since we have $$1>\epsilon\lVert z_k\rVert_{X_2}+k\lVert z_k\rVert_{X_0}>k\lVert z_k\rVert_{X_0}$$ so $$\lVert z_k\rVert_{X_0}<\frac{1}{k}$$ $\lVert z\rVert_{X_0}=\lim\limits_{k\to\infty}\lVert z_k\rVert_{X_0}=0$ implying $z=0$
but by the construction of $\{y_k\}$ we know that $\lVert z_k\rVert_{X_1}\equiv1$
which leads to contradiction.

Giotto
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