Question: given $V=C^∞(-∞,∞)$ i.e the vector space of real-valued continuous functions with continuous derivatives of all orders on $(-∞,∞)$ and $W=F(-∞,∞)$ the vector space of real-valued functions defined on $(-∞,∞)$, find a linear transformation $T:V\rightarrow W$ whose kernel is $P_3$ (the space of polynomials of degree $≤3$)
My attempt: Since $\ker(T)=\{p(x)\in V : T(p(x))=0\}=P_3$ and $\dim(P_3)=4$, my intention is if we define $T:V\rightarrow W$ by $T(f(x))=f^{(4)}(x)$ where $f^{(4)}(x)$ denotes fourth derivative of $f(x)$ at $x$, then we are done, i.e. we get $\ker T=P_3$
But, on other hand I thought, does the fourth derivative $f^{(4)}(x)=0$ imply that $f(x)$ is polynomial of degree $≤3$? How? I mean, are the only smooth functions with fourth derivative equal to $0$ polynomials of degree $≤3$?
Please help me... this is my intention about $T$ but I don't know how to find exactly what $T$ is here.
Recall that if $f'(x)$ is polynomial of degree $n$, then $f(x)$ is polynomial of degree $n+1$ by the power rule for integration. From this it follows inductively that if $f^{(k)}(x)$ is polynomial of degree $n$, then $f(x)$ is polynomial of degree $n+k$. Now, if $f^{(n)}(x)$ is identically zero, then $f^{(n-1)}(x)$ is constant and thus polynomial of degree zero, from which it follow that $f(x)$ is polynomial of degree $n-1$. Applying the case $n=4$ gives the desired result.