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Let $X=V\oplus W$ be a normed space; the title is self-explanatory: is it true that $X^*\cong V^*\oplus W^*$? What I've done is the following: I defined $F:X^*\to V^*\oplus W^*$ as $F(\phi)=(\phi|_V, \phi|_W)$. It is obvious that $F$ is well defined, linear and $1-1$. $F$ is also onto, since if $(\psi_1,\psi_2)\in V^*\oplus W^*$ we can define $\phi:X\to\mathbb{C}$ as $\phi(v+w)=\psi_1(v)+\psi_2(w)$ (since the sum is direct $\phi$ is well defined and it is obvious that $F(\phi)=(\psi_1,\psi_2)$. But the problem comes with the norms; It is $$\|F(\phi)\|=\sup_{v+w\in X, \|v+w\|=1}|\phi(v+w)|\leq\sup_{v+w\in X, \|v+w\|=1}|\phi(v)|+\sup_{v+w\in X, \|v+w\|=1}|\phi(w)|=$$ $$=\sup_{\|v\|=1}|\phi(v)|+\sup_{\|w\|=1}|\phi(w)|= \|\phi|_V\|_{V^*}+\|\phi|_W\|_{W^*}$$ but the other inequality seems impossible to me. On the other hand, $F$ seems like the only natural map that could be an isometric isomorphism. Is it not true that these two spaces are isometrically isomorphic? or am I not seeing something?

Just dropped in
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  • The category of normed spaces is usually defined so that that the morphisms are continuous linear maps. With this definition the category has finite sums and the dual of a finite sum will be isomorphic to the sum of the duals. If you require homomorphisms to be isometries, then you get a different category, which doesn't have finite sums. – Rob Arthan Feb 23 '19 at 22:33
  • If you are in A Hilbert space and the direct sum is an orthogonal direct sum then $F$ is an isometry. In general it is not. – Kavi Rama Murthy Feb 23 '19 at 23:24
  • anything unclear ? – reuns Feb 27 '19 at 00:24

1 Answers1

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I'd say it depends on the norm you chose on direct sums of normed spaces in particular on $V \oplus W$. If it is $$\|(v,w)\|_{V \oplus W}=h(\|v\|_V,\|w\|_W)$$ Where $h$ is a norm on $\mathbb{R}^2$, then let the dual norm $$h^*(a,b) = \sup_{h(x,y) = 1} |ax+by|$$ You'll have $$\|\phi\|_{(V \oplus W)^*}=\sup_{h(\|(v\|_V,\|w\|_W)=1} |\phi(v,w)|$$ $$=\sup_{h(x,y)=1}\quad\sup_{\|v\|_V = |x|,\|w\|_W = |y|} |\phi(v,0)+\phi(0,w)|=\sup_{h(x,y)=1}\quad\sup_{\|v\|_V = 1,\|w\|_W = 1} |x||\phi(v,0)|+|y||\phi(0,w)|$$ $$=h^*(\| \phi|_V\|_{V^*} , \|\phi|_W\|_{W^*})$$

For Hilbert spaces the obvious choice is $h(x,y) = \sqrt{|x|^2+|y|^2}$ so that $h^* = h$ and hence $$\|\phi\|_{(V \oplus W)^*}^2 = \| \phi|_V\|_{V^*}^2+ \|\phi_W\|_{W^*}^2$$

reuns
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