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Let $(V, \| \cdot \|)$ be a normed vector space and $M^{\perp}$ be the annihilator of $M$. If $W\subset V$ is a closed linear subspace. Prove that $R: W^{\perp} \to (V/W)^*, \ Rx^* = \tilde{x}^*$, where $\tilde{x}^*(x+ W) = x^*(x)$ is well defined.

I want to prove that when we apply $R$ on an element $x^* \in W^{\perp}$ we end up with $Rx^* \in (V/W)^*$. I know you are supposed to show the effort you made, so here is the nonsens I came up with

Let $x^* \in W^{\perp}$ and $y \in W$. Then $Rx^*(x+y) = R(x^*(x) + x^*(y)) = Rx^*(x) = \tilde{x}(x)$?

Olba12
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1 Answers1

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The elements of $W^\bot$ are the bounded linear functions $f\colon V\to k$ (where $k$ is the scalar field of $V$), such that $f(w)=0$ for all $w\in W$.

Now, we want to show that $R(f)\in (V/W)^*$, where $R(f)(x+W):=f(x)$ for any $f\in W^\bot$ and $x+W\in V/W$. So we have to verify that it is a well-defined bounded linear function $V/W\to k$.

Since the elements of $V/W$ are of the form $x+W$, we need to show first that $R(f)(x+W)=R(f)(y+W)$ whenever $x+W=y+W$. But $x+W=y+W$ iff $x-y\in W$, and then $f(x-y)=0$ because $f\in W^\bot$. Thus $$R(f)(x+W)=R(f)(y+W).$$

Now, $R(f)$ is clearly linear, because for $\alpha,\beta\in k$ and $x,y\in V$ we have $$R(f)(\alpha x+\beta y+W)=f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)=\alpha R(f)(x+W)+\beta R(f)(y+W).$$

Warning: this is not the same as showing that $R\colon W^\bot\to (V/W)^*$ is linear.

Finally, we have to show that $R(f)$ is bounded, what I'll leave to you.

Renan Mezabarba
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  • The answer clarified a lot. Well done. – Olba12 Apr 11 '17 at 19:23
  • I'm glad it helps. :) – Renan Mezabarba Apr 11 '17 at 19:23
  • What eventually made it clear for me was when I looked at $R(f)$ as a functional and not as $R$ acting on $f$. if that makes any sense. :) – Olba12 Apr 11 '17 at 19:25
  • I'll not say it doesn't make sense - who knows what really makes sense in mathematics afterall? :-) Anyway, if what you want to prove in the end of all this is the isometry between $(V/W)^$ and $W^\bot$, it may be easier to work in the other direction: considering the projection $\pi\colon V\to V/W$, we obtain an isometric isomorphism $\pi^\colon (V/W)^\to W^\bot$ by letting $\pi^(\varphi)=\varphi\circ \pi$ for any $\varphi\in (V/W)^*$. – Renan Mezabarba Apr 11 '17 at 19:29
  • It all ends up in proving an Isometric isomorphism. I will keep this in mind. :) – Olba12 Apr 11 '17 at 19:32
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    @RenanManeliMezabarba: I think the last equality should read $\alpha R(f)(x + W) + \beta R(f)(y + W)$ – el_tenedor Apr 11 '17 at 19:52
  • @el_tenedor You are absolutely right! I'll edit it now. Thank you. – Renan Mezabarba Apr 11 '17 at 19:53