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Let $X \subset Y$ be Hilbert spaces (with different norms), with continuous and dense injections. We have $$\lVert y \rVert_Y = \sup_{v \in Y, \lVert v \rVert_Y \leq 1} |(y,v)_Y|$$ Is it true that $$\lVert y \rVert_Y = \sup_{x \in X, \lVert x \rVert_X \leq 1} |(y,x)_Y|$$ I know I can take the supremum over $X$ instead but can I take the "norm less than 1" condition with the $X$ norm too? Thanks.

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  • $\begingroup$ I do not understand your question properly but what if $y\in X^\perp$. $\endgroup$ – tom Mar 7 '13 at 11:24
  • $\begingroup$ @tom maybe my edit will make that situation impossible? $\endgroup$ – michael_faber Mar 7 '13 at 11:27
  • $\begingroup$ Hilbert spaces with dense embedding? Isn't it the case then that it must be the identity? $\endgroup$ – Berci Mar 7 '13 at 11:30
  • $\begingroup$ @Berci How can that be? Consider $X = H^1$ and $Y = L^2$. They are not equal spaces but satisfy the conditions in my post? $\endgroup$ – michael_faber Mar 7 '13 at 11:36
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Counterexample:

  • $Y=\ell^2$, square summable sequences $(y_n:n\ge 1)$ with the norm $\sqrt{\sum |y_n|^2}$
  • $X=\ell^2_w$, weighted $\ell^2$ space with the norm $\sqrt{\sum n|x_n|^2}$

Clearly, $X$ is a dense subset of $Y$, since it contains all finite sequences.

Given $y\in Y$ and $x\in X$ with $\|x\|_X\le 1$, use Hölder's inequality: $$ |(y,x)_Y|= \left|\sum n^{1/2}x_n n^{-1/2}\bar y_n\right| \le \sqrt{\sum n|x_n|^2}\sqrt{\sum n^{-1}|y_n|^2 } \le \sqrt{\sum n^{-1}|y_n|^2 } $$ Clearly, the right hand side is strictly less than $\|y\|_Y$.

Considering the equality case of Hölder's inequality, one can conclude that the supremum over all eligible $x$ is indeed equal to $\sqrt{\sum n^{-1}|y_n|^2 }$. This is the norm of $y$ in $X^*$, the dual space of $X$. We have $X\subset Y\subset X^*$, a Gelfand triple.

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Since $X$ is complete, $X$ is closed in $Y$. Since $X$ is also dense, $X$ is in fact equal to $Y$. Thus your question is trivial.

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  • $\begingroup$ How do you know that $X$ is complete? $\endgroup$ – tom Mar 7 '13 at 11:32
  • $\begingroup$ How can that be? Consider $X = H^1$ and $Y = L^2$. They are not equal spaces but satisfy the conditions in my post? $\endgroup$ – michael_faber Mar 7 '13 at 11:33
  • $\begingroup$ What is $H^1{}$? $\endgroup$ – Chris Eagle Mar 7 '13 at 11:34
  • $\begingroup$ $H^1$ is the Sobolev space of $L^2$ functions with weak first order derivatives in $L^2$. $\endgroup$ – michael_faber Mar 7 '13 at 11:35
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    $\begingroup$ So when you say $X \subset Y$ are Hilbert spaces, you're allowing $X$ to have a different inner product from $Y$? How perverse. $\endgroup$ – Chris Eagle Mar 7 '13 at 11:42

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