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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.

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    $\begingroup$ 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. $\endgroup$ – AnonymousCoward Nov 20 '18 at 20:18
  • $\begingroup$ 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 $≤n$ $\endgroup$ – Akash Patalwanshi Nov 20 '18 at 20:22
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    $\begingroup$ Understand the proof of the fact I told you first, then will be able to solve your problem. $\endgroup$ – AnonymousCoward Nov 20 '18 at 20:40
  • $\begingroup$ @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? $\endgroup$ – Akash Patalwanshi Nov 20 '18 at 21:05
  • $\begingroup$ 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 ($f(x) = ax + b$). $\endgroup$ – AnonymousCoward Nov 21 '18 at 9:54
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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.

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  • $\begingroup$ Sir thanks for the answer and Sir we define degree of zero polynomial to be undefined. Is this fact can contradict your solution anywhere? Further how to prove inductive part? Please help $\endgroup$ – Akash Patalwanshi Nov 20 '18 at 21:12
  • $\begingroup$ 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 $2+2-1$. $\endgroup$ – Akash Patalwanshi Nov 20 '18 at 21:35
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    $\begingroup$ You are correct, I was a bit hasty in my reasoning; I have fixed my answer accordingly. $\endgroup$ – Reinstate Monica Nov 20 '18 at 21:40
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    $\begingroup$ In particular, I shouldn't have included the -1 term in the inductive bit. The fact that the degree of the zero polynomial is undefined doesn't affect that portion of the proof. $\endgroup$ – Reinstate Monica Nov 20 '18 at 21:41

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