Let continuous $T:X\to Y$ with $(X,||.||)$ and $(Y,||.||')$ normed spaces.Is it always true that $T(Bx)=By$? I am looking for a case where you have continuous $T:X\to Y$ where $T(Bx)$ is something else than a open (or closed) ball or even sphere ($Bx$ is the closed ball of $X$ with centre 0 and radius 1). If you have an isometry I think that is true to say $T(Bx)=By$ (where $By$ is the closed ball of $Y$ with centre 0 and radius 1) but if not I guess that you can see different things on $T(Bx)$. Any ideas?
 A: There are plenty of such examples. If $T: C[0,1] \to C[0,1]$ is defined by $Tf(x)=\int_0^{x} f(t)dt$ then the image of the closed unit ball is not a ball or  a sphere.
Note that any point in the image of the close unit ball is continuously differentiable but every ball and sphere in $C[0,1]$ contains non-differentiable functions.
A much simpler example: $(x,y) \to (2x,y)$ maps the closed unit ball in $\mathbb R^{2}$ to an ellipse.
A: Even if you impose that your map is an isometry, we need not get an open ball. Simply because the $Y$ might be "much larger" than $X$. Take for example $X=\mathbb{R}$ and $Y=\mathbb{R}^2$ and $T(x)=(x,0)$. We have $T(Bx)=T((-1,1))=(-1,1)\times \{0\}$ which is not an open ball in $\mathbb{R}^2$.
Alternatively you can consider $X=\ell^2(\mathbb{N}, \mathbb{R}) = Y$ (the space of square integrable sequences with elements in the reals). Then we can consider the "shift map" (which is also an isometry)
$$ T(x_0, x_1, x_2, \dots ) = (0, x_0, x_1, \dots). $$
Also the image here will not be an open ball (it will not even be open).
