## relation between inclusion and embedding [closed]

Assume that $X$ and $Y$ are two Banach spaces, now we have that $X$ is included in $Y$, in the sense that $\forall a\in X$, we have $a\in Y$. Then can we get that $X$ is embedded in $Y$, namely, $\forall b\in Y, \Vert b \Vert_Y \le C\cdot \Vert b \Vert_X$?

I think there is no problem for the statement of this question by Nate Eldredge.

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Is there a technical definition of 'embedding' in the context of Banach spaces? – Mariano Suárez-Alvarez Oct 21 2010 at 15:54
I don't know if this is common place, but I generally consider 'inclusion' as in set relations, and 'embedding' being a continuous inclusion (that $\|u\|_Y \leq C\|u\|_X$ for some fixed constant $C$). – Willie Wong Oct 21 2010 at 16:12
(But looking at Wikipedia, I may be the only one...) – Willie Wong Oct 21 2010 at 16:15
@Shaoming: perhaps what you want to ask is: if a map $i : X \to Y$ (thought of as "inclusion") is injective, must it be continuous? The answer is no; indeed, with the axiom of choice, one can show there exists a bijective linear map of $X$ to itself which is not continuous. – Nate Eldredge Oct 21 2010 at 20:02
@Andrew Stacey and others: It is common in the literature of this field to use the term "embedding" for "continuous injection", instead of the perhaps more logical meaning of "homeomorphism onto its image". Indeed, one often talks about one Banach space being "densely embedded" into another, which would be absurd under the latter sense. I am not sure where this usage originated; it is perhaps unfortunate but it is standard now. – Nate Eldredge Oct 21 2010 at 20:05
The inclusion of $L^\infty([0,1]$ in $L^2([0,1])$ is certainly continuous, i.e., convergence in the essential sup norm implies $L^2$ convergence, but the reverse is false. So if I understand what you are asking then I guess the answer is "no".