Why can't it map to the constant function 1? $\frac 1 x$ should work?
This is in Fraleigh Section 13, Exercise 15 for reference.
Here is the question:
Let $F$ be the set of all functions $f$ mapping $\mathbb{R}$ into $\mathbb{R}$ that have derivatives of all orders.
Is the following an isomorphism?
$\langle F, \cdot\rangle$ with $\langle F, \cdot\rangle$ where $\phi(f)(x)=x \cdot f(x)$
And the answer:
It is not an isomorphism because $\phi$ does not map $F$ onto $F$. Note that $\phi f (0) =0 \cdot f(0)=0$ . Thus there is no element of $F$ that is mapped by $\phi$ into the constant function 1.
I kind of get what it's saying but also I feel like they arbitrarily decided to write it in that order when instead they could write it like $\phi f (x)=x \cdot f(x)$ And if $f(x)=\frac 1 x$ then $x \cdot f(x) = 1$ so $\phi f(x) = 1$ even at 0, just like we wanted