A general question about inequalities

Question

I am studying Sobolev spaces, and always appear inequalities like the following $$ ||u||_{\mathcal{X}}\leq C ||u||_{\mathcal{Y}}, $$ where $\mathcal{X}=L^\infty $ and $\mathcal{Y}=W^{1,p}$ for example.

Why if I have this kind of inequalities is possible conclude that $\mathcal{Y}\subset\mathcal{X}$ with continuous injections?

Answer 1

We have the following well-known result:

Let $L:X \to Y$ be a linear map between normed vector spaces $X, Y$. Then, $L$ is continuous if and only if $L$ is Lipschitz continuous, i.e. $$\exists C > 0,\forall x \in X, \quad \|L(x)\|_Y \le C \|x\|_X.$$

Let $X, Y$ be normed vector spaces. We note $\mathcal L(X, Y)$ the set of all continuous map from $X$ to $Y$ with the usual norm, for $L \in \mathcal L(X, Y)$, $$\|L\|_\mathcal L = \sup_{\|x\|_X \le 1} \|L(x)\|_Y$$ which is well-defined since continuous map are Lipschitz continuous. Now to answer your question, if we have a canonical linear injection $\iota: X \to Y$, show that the injection is continuous is equivalent to show that there is $C > 0$ with $$\|\iota(x)\|_Y = \|x\|_Y \le C \|x\|_X.$$

Answer 2

Because if such an inequality holds for all $u\in\mathcal{Y}$ then the identity map $$I:(\mathcal{Y},\|\cdot\|_\mathcal{Y})\to(\mathcal{X},\|\cdot\|_{\mathcal{X}})$$ is an injective, bounded linear operator. For linear maps, bounded and continuous are notions that coincide.

$^*$: Just to make sure that no one gets confused with the word bounded: an operator $T:E\to F$ is called bounded when there exists a constant $c>0$ so that $\|Tx\|_F\leq c\|x\|_E$ for all $x\in E$