Suppose I have some transformation $T(f(t)) = \int_{0}^{t} f(x) dx$ from $P$ to $P$. I know this is a linear transformation, now my problem is that of proving it's isomorphic (or not, which in this case, I already know it's not).
I know that the dimensions of $P$ and $P$ are the same, so I would proceed with showing that the kernel of the transformation is $0$. I would go about doing this by showing that there is no function $f(t)$ in which the integral $\int_{0}^{t} f(x) dx = 0$, except for 0.
It's immediately obvious to me that it should be possible to create some function $f(t)$ that's negative from $0$ to $t/2$ and positive $t/2$ to $t$, or vice versa, thus showing that the kernel is not $0$. I think such a function would be $f(t) = sin(t)$, such that $T(f(2pi)) = \int_{0}^{2pi} sin(x) dx = 0$. Thus showing that $T(f(t))$ is not an isomorphism.
Is this the correct way to approach this problem?
EDIT: just realized that my approach doesn't even fall in $P$, that is to say $f(t) = sin(t)$ is not in $P$ (I think). I really don't know where to start with this problem.
For $f$ to be in the kernel, you'd need $Tf(x) = 0$ for all $x$, not just $2\pi$. In fact, you can use the fundamental theorem of calculus to show that the kernel of $T$ is $\{0\}$. Instead, you should show that $T$ is not onto (Hint: what is $Tf(0)$?)