Nam Le

A lemma of J. L. Lions

Le, Nhut Nam

Lemma statement:

Let $X$, $Y$, and $Z$ be three Banach spaces with norms $|| \cdot ||_X$, $|| \cdot ||_Y$, and $|| \cdot ||_Z$. Assume that $X \subset Y$ with compact injection and that $Y \subset Z$ with continuous injection. Prove that

$$ \forall \varepsilon > 0, \exists C_\varepsilon > 0 \text{ satisfying } || u ||_Y \leq \varepsilon || u ||_X + C _{\varepsilon}|| u ||_Z,\quad \forall u \in X $$

Applications:

  1. Prove that for every $\varepsilon > 0$ there exists $C_\varepsilon > 0$ satisfying

$$ \max_{t \in [0,1]} |u(t)| \leq \varepsilon \max_{t \in [0,1]} |u’(t)| + C_\varepsilon ||u ||_{L^1}, \quad \forall u \in C^1([0,1]). $$

  1. Pick $p > 1$. Prove that for every $\varepsilon > 0$ there exists $C = C(\varepsilon, p)$ such that

$$ || u || _{L^\infty(0,1)} \leq \varepsilon || u || _{W^{1,p}(0,1)} + C || u || _{L^1(0,1)}, \quad \forall u \in W^{1,p}(0,1). $$

Proof:

For the initial lemma, just argue by contradiction. Assume the contrary that there exists some $\varepsilon_0 > 0$ and a sequence $(u_n)_{n \in \mathbb{Z}^{+}} \subset X$ such that

$$ || u ||_Y > \varepsilon || u ||_X + C _{\varepsilon}|| u ||_Z $$

Then $u_n \ne 0, \forall n \in \mathbb{Z}^{+}$.

Let $v_n := \dfrac{u_n}{|| u_n||_X}$

Then clearly, $||v_n||_X = 1$ and we have

$$ ||v_n|| _Y > \varepsilon_0 + C _{\varepsilon_0}||v_n||_Z $$

Since $X \subset Y$ with compact injection.

Assume without loss generalization, there is $v \in Y$ such that $|| v_n - v|| _Y \rightarrow 0$ as $n \rightarrow \infty$. In particular, we have $(||v_n||) _{n \in \mathbb{Z}^{+}}$ bounded. It follows that $||v_n|| \rightarrow 0$ as $n \rightarrow \infty$.

And because $Y \subset Z$ with continuous injection, we obtain:

$$ ||v_n - v||_Z \rightarrow 0 \quad \text{as} \quad n \rightarrow \infty $$

Then $v = 0$ and $||v_n||_Y \rightarrow 0$ as $n \rightarrow \infty$

On the other hand, we also have

$$ \lim_{n \rightarrow \infty} > \varepsilon_0 + \varepsilon_0\lim_{n \rightarrow \infty}||v_n||_Z $$

Consequently,

$$ 0 > \varepsilon_0 > 0 $$ which is a contradiction. The two application are more or less immediate after using the given lemma. The proof is completed.

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