-1
$\begingroup$

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.

Improve this question
$\endgroup$
11
  • $\begingroup$ Is there a technical definition of 'embedding' in the context of Banach spaces? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 21, 2010 at 15:54
  • $\begingroup$ 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 $C$). $\endgroup$
    – Willie Wong
    Commented Oct 21, 2010 at 16:12
  • $\begingroup$ (But looking at Wikipedia, I may be the only one...) $\endgroup$
    – Willie Wong
    Commented Oct 21, 2010 at 16:15
  • 2
    $\begingroup$ @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. $\endgroup$
    – Nate Eldredge
    Commented Oct 21, 2010 at 20:02
  • 2
    $\begingroup$ @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. $\endgroup$
    – Nate Eldredge
    Commented Oct 21, 2010 at 20:05

1 Answer 1

Reset to default
2
$\begingroup$

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".

Improve this answer
$\endgroup$
2
  • $\begingroup$ but I think we have that, L^(inf) is embedded in L^2, sorry for not using language of Latex $\endgroup$
    – Shaoming Guo
    Commented Oct 21, 2010 at 19:05
  • $\begingroup$ In customary terminology, an injective map is called an immersion if it is continuous, but it is only called an embedding if it is bi-continuous. $\endgroup$
    – Dick Palais
    Commented Oct 21, 2010 at 19:32

Not the answer you're looking for? Browse other questions tagged .