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I'm working on the extra credit for my Calculus 1 class and the last problem is a proof. We have done proofs before, but I'm unsure of how to approach this problem. Any help would be much appreciated, I'm just not entirely sure how to start. Specifically, showing how the continuous function defined by the problem is to taken into account. I understand that it defines the bounds for the integrals, but beyond that I am unsure of how they are affected. My intuition tells me that it implies the use of a specific theorem, but no specific theorem is apparent to me from simple observation. I've found evidence that Holder's Inequality would be helpful, but I'm not certain as to how to apply this. The problem is as follows:

Show that for any continuous function $f:[-1,1]\to\mathbb R$ we have that:

$$2\int_{-1}^{1} f^2(x) dx \ge \left(\int_{-1}^{1}f(x)dx\right)^2+3\left(\int_{-1}^{1}xf(x)dx\right)^2$$

I've tried multiple approaches essentially yielding nothing though. My understanding of how to manipulate integrals in a general manner is weak at best. I did get a hint from my TA that said the proof is almost a page long and uses results beyond Calculus 1, so technically I shouldn't even be able to do this proof, but it is worth a significant amount of my grade and I could really use all the help I can get. We actually only started integrals about 3 weeks ago. Not to mention it has now started to bother me that I can't solve it despite spending so much time. The extra credit is due Friday, but my Final is tomorrow, so I won't be able to spend much more time on it.

Thanks for any help you guys can provide.

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  • $\begingroup$ I have modified the title. $\endgroup$ – Theo May 14 '15 at 1:44
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For brevity, write $\int F $ for $ \int_{-1}^{1} F(x) \, dx $.

Let $\int f = A$. Then $$ \int \left( f-\tfrac{1}{2}A \right) = \int f - A = 0 $$ Now use Cauchy–Schwarz: $$ 3\left( \int x \left(f-\tfrac{1}{2}A \right) \right)^2 \leqslant 3\left( \int x^2 \right) \left( \int \left(f-\tfrac{1}{2}A \right)^2 \right) = 2 \left( \int f^2 - A\int f + \frac{A^2}{4} \int 1 \right) \\ = 2\left( \int f^2 - A^2 +\frac{A^2}{2} \right) = 2\int f^2 -A^2 $$ Also, we have $$ 3\left( \int x(f-A) \right)^2 = 3\left(\int xf\right)^2, $$ because $\int x = 0 $, and so we are done, after rearranging.

(Actually, you don't need $f$ to be continuous, just that all three of the integrals exist.)


Now, I shall do you one better: there is a set of polynomials $P_n(x)$, where $P_n$ is of degree $n$, such that you can find a similar inequality in the same way: for every positive integer $N$, if $\int x^n f$ exists for $0 \leqslant n \leqslant N$, there is an inequality $$ 2\int f^2 \geqslant \sum_{n=0}^N (2n+1)\left( \int P_n f \right)^2 $$ If we take $N=1$, then this is for $P_0(x)=1$, $P_1(x)=x$. The key to the proof is the orthogonality of the $P_n$: we have $$ \int P_n P_m = \begin{cases} \frac{2}{2n+1} & n = m \\ 0 & n \neq m \end{cases}. $$ We can find unique $P_n$ with this property, given only that $P_n$ has degree $n$. These are called the Legendre polynomials, probably the most basic example of orthogonal polynomials.

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  • $\begingroup$ Wow, it would seem my professor was having some fun at the expense of my class. This isn't even an honors course and I do not attend a school with a strong math focus, so I'm not quite sure what he was expecting from us beyond some completely nonsensical scribbles. It will be very clear that this proof is not of my own making, at the very least, because I will state as much, but he cannot fault me for attempting the proof and then seeking help. Thank you very much for your help @Chappers and best of luck with your studies. $\endgroup$ – Theo May 14 '15 at 2:51
  • $\begingroup$ I worry about how the person who thought it took almost a page to prove was going to prove it: the tricky part here is having the idea (which you should know did not come instantly: I had to try a few things before I worked out what the correct way to start was). Once you have the right idea, the rest is just algebra. $\endgroup$ – Chappers May 14 '15 at 3:10
  • $\begingroup$ Well my TA is honestly a very adept mathematician and was published as an undergrad, so personally I do not doubt the efficacy of his proof. Based on the handwritten solutions he normally provides for tests I would really just assume the length is due to a combination of his handwriting (large) and his likely writing the proof in a very formal manner. I mean I understand some of the concepts applied here, but very informally and only after about an hour of reading wiki's. Really the algebra involving integrals is new to me as well, so I can't even verify the proof myself. $\endgroup$ – Theo May 14 '15 at 3:49
  • $\begingroup$ Hi again @Chappers, I've been working over the proof for a bit now after taking my final and just have a couple questions about the setup if you're not too busy. What is the significance of the Integral F at the beginning vs the Integral f used within the problem? I'm just not sure what the change in capitalization indicates. The second is how A is defined and then used within integrals. Am I to assume we're integrating an integral? Or is A just used to represent an integral we manipulate within and around the other integrals? I hope my questions were clear. Thanks again! $\endgroup$ – Theo May 14 '15 at 20:54
  • $\begingroup$ @Chapers Ah, I see now that the A represents a definite integral that will in less general terms be a constant. $\endgroup$ – Theo May 14 '15 at 22:02

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