# Norm of element of Hilbert space as supremum over dense subspace?

Suppose $H_1 \subset H_2$ are both Hilbert spaces with different inner products $(\cdot,\cdot)_{H_1}$ and $(\cdot,\cdot)_{H_2}$. Suppose also that $H_1$ is dense in $H_2$ and that the inclusion is continuous (as in case of Gelfand triple).

Now, $$\lVert h_2 \rVert_{H_2} = \sup_{h \in H_2, \lVert h \rVert_{H_2} = 1}|(h_2, h)|$$ is by definition almost.

Question: Is it true that $$\lVert h_2 \rVert_{H_2} = \sup_{h \in H_1, \lVert h \rVert_{H_1} = 1}|(h_2, h)|?$$

That is, can I take the supremum over a dense subspace?

(In a previous thread Norm of linear functional; can we take supremum over dense subset?, related to this question, I did not clarify the situation enough).

• Your description of the meaning of the question is misleading: for example, as in the inclusion $H^1\subset L^2$ that arises in the "Gelfand triple" stories, for example Levi-Sobolev space on a circle, the unit ball in $H^1$ is by far not dense in the unit ball in $L^2$. Often these inclusions are compact, so the image of the finer topology's unit ball much be much smaller than the unit ball in the coarser. – paul garrett Mar 11 '13 at 20:30
• @paulgarrett I don't understand, I thought $H^1(M) \subset L^2(M)$is always compact and dense for compact manifolds $M$. – maximumtag Mar 11 '13 at 22:34
• The map is compact, yes. The image of the unit ball from $H^1$ is pre-compact in $L^2$ in such cases, yes, and the image of $H^1$ in $L^2$ is dense, yes. But the sup of $\langle x,y\rangle_{L^2}$ with $|x|_{H^1}\le 1$ will be smaller than $|y|_{L^2}$ for typical $y$ in $L^2$. Nothing pathological or tricky about this, in fact. – paul garrett Mar 12 '13 at 0:04
• What is the inner product you are using inside the supremum? You might get better answers if you provide more information on the background of the question. – timur Mar 13 '13 at 3:18

It is false. For example, let $H_1=H_2$ as sets and let $(\cdot,\cdot)_{H_2}=c^2(\cdot,\cdot)_{H_1}$ for some $c>0$. Then $$\sup_{h\in H_1,\|h\|_{H_1}=1}|(h_2,h)_{H_2}|=\sup_{h\in H_2,\|h\|_{H_2}=c}|(h_2,h)_{H_2}|=c\|h_2\|_{H_2}.$$
• Thanks. But in your example the norms are equivalent. If this is not true do you know if my desired result will hold? Eg. $H^1(\Omega) \subset L^2(\Omega).$ Indeed I don't need equality, a $\leq$ will do. – maximumtag Mar 11 '13 at 21:56
• @maximumtag: It still does not hold, even if for $\le$, because if your result holds for some $(,)_{H_1}$, you can replace it with another equivalent inner product as given in my answer. – 23rd Mar 12 '13 at 2:40