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Can you please give me a simple example of linear continuous mapping from non separable normed space $X$ onto separable normed space $Y$.

Thanks a lot.

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Consider maps from $\ell_\infty\oplus_1\ell_1$ to $\ell_1$. – David Mitra Jul 29 '12 at 17:44
up vote 1 down vote accepted

Let $X$ be a non-separable normed space.

If $Y$ is finite dimensional, then for every non-separable space $X$ the answer to your question is positive. Since $Y$ is finite dimensional it is, of course, separable. Consider arbitrary closed subspace $X_0\subset X$ with codimension equal to $\mathrm{dim}(Y)$. Since this dimensions are finite you have a lot of ismorphisms between $X/X_0$ and $Y$. Lets take one $I:X/X_0\to Y$. Consider standard projection $\pi :X\to X/X_0$, then the desired map is $T=I\circ \pi$. It is linear continuous and surjective as composition of continuous surjective linear maps.

If $Y$ is separable and infinite dimensional then for some non-separable spaces $X$ the answer to your question is positive. The easiest example (provided by David Mitra) is the following: take $X=\ell_\infty\oplus_1 Y$ and consider projection $$ \pi:\ell_\infty\oplus_1 Y\to Y:(z,y)\mapsto y $$

In general if $Y$ is infinite dimensional and separable, it is not know whether for arbitrary non-seprable $X$ there exist surjective operator $T:X\to Y$. In fact this question is related to the following well known open "separable quotients problem": $$ \text{ Does every Banach space have separable quotient } $$ $$ \text{ by closed subspace of infinite codimension? } $$ There are wide classes of non-separable Banach spaces whose quotients are separable. For details see this answer on mathoverflow.

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Maybe I misunderstand what is meant. Finite codimension does not imply closed. E.g., if you take a discontinuous linear operator onto a finite dimensional space, its kernel is finite codimensional and not closed. However, if the range of a continuous linear operator is finite codimensional then it is closed. – Jonas Meyer Jul 30 '12 at 17:33
@JonasMeyer, so I'll just require closedness. Is it true that for each $n\in\mathbb{N}$ there exist closed subspaces with codimension $n$? – Norbert Jul 30 '12 at 17:45
I guess you can take linearly independent $\varphi_1,\ldots,\varphi_n\in X^*$ (e.g. using Hahn-Banach), define $T:X\to\mathbb C^n$ by $T(x)=(\varphi_1(x),\ldots,\varphi_n(x))$, and let $X_0=\ker T$. – Jonas Meyer Jul 30 '12 at 17:50
So, my argument works... Thanks, Jonas you always help me so much! – Norbert Jul 30 '12 at 18:04
Why not state it positively: does every Banach space have a separable infinite-dimensional quotient (by a closed subspace)? – t.b. Jul 30 '12 at 18:28

Let $X=\ell^\infty(\mathbb{N})$, $Y=\mathbb{C}$. Define $\varphi:X\to Y$ by $$ \varphi(x)=x(1). $$

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Take any bounded non-zero functional from a non-separable normed space.

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