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Let $V \subset H$ where $H$ is Hilbert space. Let $T:H^* \to V^*$ be the canonical map that restricts the domain of a functional in $H$ so that it's a functional in $V$.

How do I show that $$\lVert T\varphi \rVert_{V^*} \leq C\lVert \varphi \rVert_{H^*}$$ if $\lVert v \rVert_H \leq C\lVert v \rVert_V$? I can't get it out...

Also, how to show that

$Range(T)$ is dense in $V^*$ if $V$ is reflexive?

No idea where to start with this. I only know that reflexive means $V \equiv V^{**}$.

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This is a rather bizarre question, since Hilbert spaces are all reflexive. –  Chris Eagle Oct 23 '12 at 20:52
    
@ChrisEagle We can have $H=\ell^2(\Bbb C)$ and $V=\ell^1(\Bbb C)$, so $V$ is not necessarily a Hilbert space, and not necessarily reflexive. –  Davide Giraudo Oct 24 '12 at 11:35
    
@Davide Giraudo: when I made my comment, the question stated that $V$ is a Hilbert space. –  Chris Eagle Oct 24 '12 at 11:48
    
@ChrisEagle Yes, I remember now. –  Davide Giraudo Oct 24 '12 at 12:01

2 Answers 2

up vote 2 down vote accepted

$\def\norm#1{\left\|#1\right\|}\def\abs#1{\left|#1\right|}$We have for $\phi \in H^*$ \begin{align*} \norm{T\phi}_{V^*} &= \sup_{\norm v_V = 1} \abs{(T\phi)(v)}\\ &= \sup_{\norm v_V = 1} \abs{\phi(v)}\\ &\le \sup_{\norm h_H \le C} \abs{\phi(h)} & \text{as }\{v \in V \mid \norm v_V = 1\} \subseteq \{h \in H \mid \norm h_H \le C\}\\ &\le \sup_{\norm h_H \le C} \norm{\phi}_{H^*} \norm h_H\\ &= C \norm{\phi}_{H^*} \end{align*} Note that we didn't use the Hilbert space property of $H$ or $V$ here, this works for general Banach spaces, which is also the context in which the second question makes sense, as Hilbert spaces are always reflexive. So let $V$ be a reflexive Banach space and $W := \overline{\operatorname{ran} T} \subseteq V^*$. We will show that every $\psi \in V^{**}$ with $\psi|_W = 0$ is zero on $V^*$. By Hahn-Banach, this suffices. So let $\psi \in V^{**}$ with $\psi|_W = 0$ be given, as $V$ is reflexive, there is a $v \in V$ with \[ \psi(v^* ) = v^* (v), \quad \text{all } v^* \in V^*. \] We now have for each $\phi\in H^*$ that \[ 0 = \psi(T\phi) = (T\phi)(v) = \phi(v) \] which is, by Hahn-Banach, only possible for $v = 0$ (If $v \ne 0$, we could extend the functional $\mathbb K\cdot v \to \mathbb K$, $\lambda v \mapsto \lambda\cdot \|v\|$ continously to a $\phi \in H^*$, then $\phi(v) = \|v\| \ne 0$). So $\psi = 0$ and $W = V^*$, as wished.

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Thanks, please can you expand on the last sentence. I think my H-B is different to yours.. –  hopo2 Oct 24 '12 at 17:12
    
@hopo2 I added something. –  martini Oct 24 '12 at 20:57

Let $v\in V$. As $(T\varphi)(v)=\varphi(v)$ if $\varphi\in H^*$, we have $$|T(\varphi)(v)|=|\varphi(v)|\leq \lVert \varphi \rVert_H\cdot\lVert v\rVert_H\leq C\lVert \varphi \rVert_H\cdot\lVert v\rVert_V,$$ which gives for $v\neq 0$: $$\frac{|T(\varphi)(v)|}{\lVert v\rVert_V}\leq C\lVert \varphi \rVert_H,$$ which gives, after having taken the supremum, the wanted result.

For the second question, let $L$ a linear functional on $V^*$ (hence an element of $V^{**}$) which vanishes on $T(V^*)$. Let $J\colon V\to V^{**}$ the canonical inclusion; here it's surjective, hence $L=J(l)$ for some $l\in V$. We have $$\langle L,T\varphi\rangle_{V^{**},V^*}=\langle J(l),T\varphi\rangle_{V^{**},V^*}=\langle T\varphi,l\rangle_{V^*,V}=0.$$ As $l\in V$, we have for all $\varphi\in H^*$, $\varphi(l)=0$ so $l=0$, and we conclude by Hahn-Banach theorem.

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