3

I know this uses an integration by parts trick, but I seem to be doing something very wrong. I'm trying to show that given f is a real, continuouously differentiable function on [a,b], f(a)=f(b)=0 and $\int_a^{b} f(x)^2 dx =1, \int_a^{b} xf(x)f'(x) dx = \frac{-1}{2}$ and $\int_a^{b}(f'(x))^2 dx \int_a^{b} x^2(fx)^2 dx > 1/4$

For the first part, I'm trying to use integration by parts on $\int_a^{b} xf(x)f'(x) dx$.

So we let u=x, and $v=f(x)f'(x)$. Then integrating v we get $dv = \frac{1}{2} f^2(x)$. And so we have that:

$\int_a^{b} xf(x)f'(x) dx = x\frac{1}{2}f^2(x) |_a^{b} - \int_a^{b} \frac{1}{2}f^2 dx = b\frac{1}{2} \cdot (1) - (a)\frac{1}{2} \cdot 1 - 0?$

Where am I going wrong?

Ben Grossmann
  • 225,327
Nikitau
  • 1,353
  • 1
    This seems obviously wrong. Say it is true for $f(x)$. then let $g(x)=cf(x)$. We still have $g(a)=0=g(b)$ but now the integral is multiplied by $c^2$. – lulu Dec 20 '16 at 19:24
  • As a concrete example, $\int_0^{\pi}x\sin(x)\cos(x)dx = -\frac {\pi}4$. – lulu Dec 20 '16 at 19:25
  • @lulu Maybe I made a typo that I'm not seeing, but there is a similar question here http://math.stackexchange.com/questions/1563223/if-fa-fb-0-show-that-int-ab-xfxfxdx-frac-12-intfx2dx for the first part, but I still can't seem to figure it out. – Nikitau Dec 20 '16 at 19:25
  • for the second part - see f(x) = 0 – kotomord Dec 20 '16 at 19:26
  • That question is radically different....the integral there still depends on $f(x)$. – lulu Dec 20 '16 at 19:27
  • Wait, are you assuming that $\int_a^b \left( f(x) \right)^2dx =1 $? That assumption seems to just creep in at some point. If so, then the claim is trivial...by parts, as in your calculation. – lulu Dec 20 '16 at 19:29
  • @lulu If I understand you correctly, my part of the question assumes that the integral left behind $\int_a^{b} f^2(x) = 1$. It is stated in the body of the question – Nikitau Dec 20 '16 at 19:29
  • Yeah, I just saw that. But in that case your argument by parts is complete. – lulu Dec 20 '16 at 19:29

1 Answers1

2

You should have $$ \int_a^{b} xf(x)f'(x) dx = \left. x\frac{1}{2}f^2(x) \right|_a^{b} - \int_a^{b} \frac{1}{2}f^2 dx = 0 - \frac 12 $$ since $f^2(a) = f^2(b) = 0$. I have no idea how you ended up with $b\frac{1}{2} \cdot (1) - (a)\frac{1}{2} \cdot 1 - 0$.

From there, think Cauchy Schwarz.

Ben Grossmann
  • 225,327