function from non separable normed space $X$ onto separable normed space $Y$. 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.
 A: Let $X=\ell^\infty(\mathbb{N})$, $Y=\mathbb{C}$. Define $\varphi:X\to Y$ by 
$$
\varphi(x)=x(1).
$$
A: Take any bounded non-zero functional from a non-separable normed space. 
A: 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.
