## Ishaan94 3 years ago 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 $\int _0^{2\pi} f(x) \cos x dx \ge 0$

1. Ishaan94

@myininaya @amistre64 How would you prove it?

2. Ishaan94

@Mr.Math

3. Mr.Math

$\int_0^{2\pi}f(x)\cos(x)dx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin(x)dx=-\int_0^{2\pi}f'(x)\sin(x)dx$ $=f'(x)\cos(x)|_0^{2\pi} +\int_0^{2\pi}f''(x)\cos(x)dx=f'(2\pi)-f'(0)+\int_0^{2\pi}f''(x)\cos(x)dx.$ 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

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

f"(x) must be constant, right?

6. Mr.Math

Why?

7. Mr.Math

It doesn't have to be constant.

8. myininaya

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

9. Mr.Math

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

10. Ishaan94

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

11. Mr.Math

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

I didn't defined it as a constant.

13. Mr.Math

define*

14. Mr.Math

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

Am I making sense Ishaan?

16. Ishaan94

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

17. Mr.Math

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

18. Ishaan94

Do you think I should ping Zarkon?

19. Mr.Math

I think I have it. One minute.

20. Mr.Math

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

21. Ishaan94

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

Correct! I just noticed that :)

23. Mr.Math

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

24. Mr.Math

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

25. Ishaan94

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

26. satellite73

i am thinking parts

27. satellite73

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

28. Mr.Math

Yes.

29. Ishaan94

yeah

30. Mr.Math

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

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

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

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

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

It is a given in the problem

36. Mr.Math

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

ooooooooh ok i see i misinterpreted the last step

38. Mr.Math

@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

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

40. Mr.Math

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 $\int_0^{2\pi}f''(x)\cos(x)dx \le \int_0^{2\pi}f''(x)dx=f'(2\pi)-f'(0). ■$