3
$\begingroup$

So suppose $f$ is a generally complex, continuous function on $[0,1]$ and $P$ is a polynomial defined on the real numbers. Evidently $$ \int_0^1 f(t)P(t - x)\,dt $$ is a polynomial in $x$ but that this is the case is not obvious to me. How might one go about proving this? Note that $f$ is continuous and so is integrable but it may not have a nice anti-derivative.

$\endgroup$
3
  • 1
    $\begingroup$ The intuitive answer I would give a student would be that the $t$ terms disappear. Remember that definite integrals with respect to a given variable $\theta$ turn all instances of $\theta$ into constants, so all that is left is a polynomial in $x$ (and integration of a polynomial yields another polynomial). Of course, this explanation doesn't quite explain that the $t-x$ within the polynomial function leaves the function a polynomial once integrated, though this too can be reasoned by a similar thought process and the fact that integrating a polynomial yields another polynomial $\endgroup$ Jun 29 '16 at 2:33
  • $\begingroup$ Additionally, the proof by Alozizio below should help in piecing together the above logic and provides a more rigorous foundation for the claim that $x \to t-x$ doesn't change the fact we get a polynomial in $x$ $\endgroup$ Jun 29 '16 at 2:36
  • $\begingroup$ @BrevanEllefsen: The intuitive answer you gave is rigorous and is almost devoid of mathematical symbols. I like such arguments because they are easier to understand and do not compromise on rigor. $\endgroup$ Jun 29 '16 at 6:49
5
$\begingroup$

$P(X)=a_0+a_1X+\cdots+a_nX^n.$

$P(t-X)=a_0+a_1(t-X)+\cdots+a_n(t-X)^n=a_0+a_1t-a_1X+a_2t^2+2a_2tX+a_2X^2+\cdots$

$\implies f(t)P(t-X)=f(t)a_0+tf(t)-a_1X+t^2f(t)a_2+2a_2tf(t)X+f(t)a_2X^2+\cdots $

The above is meant for you to visualize what is happening. When you integrate with respect to $t$, everything alongside the $X$'s will turn to constants.

$\endgroup$
5
$\begingroup$

Alternatively, show that $Q(x):=\int_0^1\,f(t)\,P(t-x)\,\text{d}t$ satisfies $\frac{\text{d}^k}{\text{d}x^k}\,Q(x)=0$ for some integer $k>0$. As $P$ is a polynomial, $\frac{\partial^k}{\partial x^k}\,P(t-x)=0$ if $k:=1+\deg(P)$. Thus, the Leibniz Integral Rule gives $$\frac{\text{d}^k}{\text{d}x^k}\,Q(x)=\int_0^1\,f(t)\,\frac{\partial^k}{\partial x^k}\,P(t-x)\,\text{d}t=0\,.$$ This result holds for any integrable function $f:[0,1]\to\mathbb{C}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.