## Ishaan94 Group Title Let $$f(x)$$ be a continuous function, whose first and second derivatives are continuous on $$[0,2\pi]$$ and $$f"(x) \ge 0 \:\: \forall \:x \in [0,2\pi]$$. Show that $\int _0^{2\pi} f(x) \cos x dx \ge 0$ 2 years ago 2 years ago

1. Ishaan94 Group Title

@myininaya @amistre64 How would you prove it?

2. Ishaan94 Group Title

@Mr.Math

3. Mr.Math Group Title

$\int_0^{2\pi}f(x)\cos(x)dx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin(x)dx=-\int_0^{2\pi}f'(x)\sin(x)dx$ $=f'(x)\cos(x)|_0^{2\pi} +\int_0^{2\pi}f''(x)\cos(x)dx=f'(2\pi)-f'(0)+\int_0^{2\pi}f''(x)\cos(x)dx.$ We can see here that $$f'(2\pi)-f'(0)\ge 0$$, since $$f'(x)$$ is an increasing function. So we're left to show that $$\int_0^{2\pi}f''(x)\cos(x)dx\ge 0$$ knowing that $$f''(x)\ge 0$$.

4. Mr.Math Group Title

Let $$g(x)=\int f''(x)\cos(x)dx$$. The question here is how can we show that $$g(2\pi)\ge g(0)$$? We know that $$g(x)$$ is increasing on $$(0,\frac{\pi}{2})∪(\frac{3\pi}{2},2\pi)$$, and decreasing on $$(\frac{\pi}{2}, \frac{3\pi}{2}).$$

5. Ishaan94 Group Title

f"(x) must be constant, right?

6. Mr.Math Group Title

Why?

7. Mr.Math Group Title

It doesn't have to be constant.

8. myininaya Group Title

mr.math do you mean g is positive on those intervals...and g is negative on that interval...

9. Mr.Math Group Title

I meant $$g'$$ is positive on those interval and negative on that interval.

10. Ishaan94 Group Title

I know, I just wanted it to be constant I mean not any obvious reason, just like that.

11. Mr.Math Group Title

Note that $$g'(x)=f''(x)\cos(x)$$, where $$f''(x)\ge 0$$. So the sign of $$g'(x)$$ (that determines whether g is increasing or decreasing) can be determined by $$\cos(x)$$.

12. Mr.Math Group Title

I didn't defined it as a constant.

13. Mr.Math Group Title

define*

14. Mr.Math Group Title

I don't understand why you want $$f''(x)$$ to be a constant. If it was then the solution would be obvious since g(2pi)=g(0), and the integral will be 0.

15. Mr.Math Group Title

Am I making sense Ishaan?

16. Ishaan94 Group Title

Of course, you're... it's me, who didn't

17. Mr.Math Group Title

This problem is not difficult, but I'm missing something. What am I missing?!

18. Ishaan94 Group Title

Do you think I should ping Zarkon?

19. Mr.Math Group Title

I think I have it. One minute.

20. Mr.Math Group Title

You can ping him though, but he should not answer until I finish :P

21. Ishaan94 Group Title

If I'm not mistaken,$- \int_0^{2\pi} f'(x) \sin x dx = f'(x) \cos x|_0^{2\pi} - \int _0^{2\pi} f''(x) \cos xdx$

22. Mr.Math Group Title

Correct! I just noticed that :)

23. Mr.Math Group Title

So we have to show that $$f'(2\pi)+g(0)\ge g(2\pi)+f'(0).$$

24. Mr.Math Group Title

I'm getting tired. @Zarkon should come from his planet now.

25. Ishaan94 Group Title

Zarkon's offline :(... @satellite73 @eseidl

26. satellite73 Group Title

i am thinking parts

27. satellite73 Group Title

oh i should pay attention. looks like mr.math did parts right?

28. Mr.Math Group Title

Yes.

29. Ishaan94 Group Title

yeah

30. Mr.Math Group Title

I think I have it this time. By the squeeze theorem and property of inequality in integrals we have: $-\int_0^{2\pi}f''(x) dx\le \int_0^{2\pi}f''(x)\cos(x)dx\le \int_0^{2\pi}f''(x)dx.$ Thus $$g(2\pi)-g(0)\le f'(2\pi)-f'(0)$$ as required.

31. satellite73 Group Title

that was nice. i was trying to come up with a counter example, and example of a positive function where this integral would be negative. can't seem to do it.

32. satellite73 Group Title

still not sure why i cannot though. why can't it be postive, but larger on the interval $[\frac{\pi}{2},\frac{3\pi}{2}]$ then on the rest? wouldn't that make the integral negative?

33. Mr.Math Group Title

After this proof, I am sure you can't find such an example :P @Ishaan, I will go for few minutes and be back to summarize my solution in one post, so it can be read and understood easily.

34. satellite73 Group Title

i believe you, and understand the first part. but it looks like the second part is hinging on the fact that if $f''>0$ then $\int f''(x)\cos(x)dx\geq 0$ and i am not sure exactly why that has to be

35. eseidl Group Title

It is a given in the problem

36. Mr.Math Group Title

Well, what I proved is that a continuous function $$f(x)$$ that has a positive second derivative has the property: $\int_0^{2\pi} f(x)\cos(x)dx\ge 0.$

37. satellite73 Group Title

ooooooooh ok i see i misinterpreted the last step

38. Mr.Math Group Title

@satellite I first said that, but I don't need that statement anymore (I don't know if it's true or not). Read the last two posts.

39. satellite73 Group Title

yes, sorry i see what you wrote, i simply read it wrong.

40. Mr.Math Group Title

We can use integration by parts and write $\int_0^{2\pi}f(x)\cos(x)dx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin(x)dx=-\int_0^{2\pi}f'(x)\sin(x)dx$ $=f'(x)\cos(x)|_0^{2\pi} +\int_0^{2\pi}f''(x)\cos(x)dx=f'(2\pi)-f'(0)-\int_0^{2\pi}f''(x)\cos(x)dx.$ From here it's sufficient to show that $$f'(2\pi)-f'(0)\ge\int_0^{2\pi}f''(x)\cos(x)dx.$$ We know that $$\cos(x)\le 1 \implies f''(x)\cos(x)\le f''(x)$$. So $\int_0^{2\pi}f''(x)\cos(x)dx \le \int_0^{2\pi}f''(x)dx=f'(2\pi)-f'(0). ■$