# Find linear transformation whose kernel is given

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.

• Hint: Here is a fact that can be used to answer a simpler version of your question. Let $f$ continuously differentiable. Then $f' = 0$ (as functions) iff $f$ is a constant function. – AnonymousCoward Nov 20 '18 at 20:18
• Sir, thanks for reply. Can you tell me then how can we prove " if $f$ is infinitely continuously differentiable then $f^{m}=0$ iff $f$ is polynomial of degree $≤n$ – Akash Patalwanshi Nov 20 '18 at 20:22
• Understand the proof of the fact I told you first, then will be able to solve your problem. – AnonymousCoward Nov 20 '18 at 20:40
• @AnonymousCoward sir, using mean value theorem we can easily prove that $f'=0$ iff $f$ is constant. Sir how does it help me to prove the advanced version? – Akash Patalwanshi Nov 20 '18 at 21:05
• The next step is to use the same idea to prove that for $f$ continuously differentiable: $f'$ is a constant function iff $f$ is a linear function ($f(x) = ax + b$). – AnonymousCoward Nov 21 '18 at 9:54

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.
• Sir, I think there is some mistake in your solution? take $f(x)=x^4$ then $f'(x)=4x^3$, $f"(x)=12x^2$, $f^{3}(x)=24x$....Now, clearly here $f"$ is polynomial of degree $2$ but $f(x)$ is not polynomial of degree $2+2-1$. – Akash Patalwanshi Nov 20 '18 at 21:35