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I am looking at problems from a released Fall 14 mock exam. The question in particular is number 2:

Let $f$ be a continuous function in $[0,1]$ satisfying the condition:

$$ \int_x^1 f(t) dt \geq \frac{1-x^2}{2}$$ for $x \in [0,1]$

Prove that:

$$\int_0^1 |f(x)|^2 dx \geq \int_0^1 xf(x)dx$$

This is what I have come up with so far:

First off we can "evaluate" the integral from $0$ to $1$:

$$\int_0^1 f(t) dt \geq \frac{1-0^2}{2} = \frac{1}{2}$$

Next we know from the Cauchy-Schwartz inequality:

$$\left|\int_0^1 f(x) dx\right|^2 \leq \int_0^1 |f(x)|^2 dx$$

So:

$$\int_0^1 |f(x)|^2 dx \geq \frac{1}{4}$$

Now for the other equation. I used integration by parts:

$$\int_0^1 xf(x)dx = xF(x) - \int_0^1 F(x) dx $$

$$\int_0^1 xf(x) dx \geq 1 \cdot \frac{1}{2} - \int_0^1 F(x) dx $$

But notice that:

$$F(x)|_0^1 \geq \frac{1}{2}$$

So:

$$\int_0^1 xf(x) dx \geq \frac{1}{2} - \frac{1}{2}$$

$$\int_0^1 xf(x) dx \geq 0$$

Which seems to be right so far (I could of course be wrong). I don't have enough info to close anything out, but it seems to be pointing in the right direction. Any further hints or corrections would be greatly appreciated.

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Dair
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  • Your last one but step is incorrect because the inequality for the second term is going the other way. – Aravind Sep 29 '15 at 06:21
  • Oh, I see silly me. The negative sign should flip the inequality. I still believe it results in the same final step though unless I am misunderstanding? – Dair Sep 29 '15 at 06:24
  • Rewrite the first inequality as $\int_{x} ^ {1} [f(t)-t] dt \geq 0$. Perhaps from this it is possible to conclude that $f(x) \geq 0$ for all $x$. – Aravind Sep 29 '15 at 06:41
  • You should probably apply Cauchy-Schwarz to the pair of functions $f(x)$ and $x$. – Aravind Sep 29 '15 at 06:56
  • Ok, so if I understand this correctly, then: $$\int_x^1 [f(t) - t] dt \geq 0$$ $$\int_x^1 f(t) dt - \int_x^1 t dt \geq 0$$ $$\int_x^1 f(t) dt \geq \int_x^1 t dt$$ So from this: $$\left| \int_0^1 f(x) \right|^2 \geq \int_0^1 xf(x) dx$$ and from the above Cauchy-Schwarz inequality one can conclude: $$\int_0^1 |f(x)|^2 dx \geq \int_0^1 x f(x) dx$$ – Dair Sep 29 '15 at 08:37
  • We cannot conclude thus. I meant the inequality $(\int_{o}^{1}|f(x)^2|dx )^{\frac{1}{2}} ( \int_{0}^{1} x^2dx)^{\frac{1}{2}} \geq \int_{0}^{1}xf(x)dx$, but this does not seem to yield the desired conclusion. – Aravind Sep 29 '15 at 08:55
  • @Aravind: Sorry, but do you mind clarifying why we cannot conclude thusly? – Dair Sep 29 '15 at 09:03
  • you should try and plug in various affine functions for $f$ and see what hapens. Try to find a case where the goal is an equality. This can guide you and prevent you from trying to prove wrong things like $\frac 14 \ge \int_0^1 tf(t) dt$ and it also tells you that you do need to use the full hypothesis given to you and not just $\int_0^1 f(t)dt \ge \frac 12$. – mercio Sep 29 '15 at 11:31
  • @mercio: Plugging in the linear transformation $t$ gives an equality for the first statement. So then looking at the comment above you can use this fact to conclude: $\int_x^1 f(t) dt \geq \int_x^1 t dt$. Therefore: $$\left|\int_0^1 f(x)dx \right|^2 \geq \int_0^1 x dx \cdot \int_0^1 x dx$$ and similarly from the Cauchy-Schwarz inequality: $$\int_0^1 |f(x)|^2 dx = \int_0^1 x f(x) dx$$ – Dair Sep 29 '15 at 21:51

2 Answers2

1

Define $g=f^{+}=\max(f,0)$, $h=f^-=-\min(f,0)$, we have $f=g-h$, $|f|=g+h$. $g\geq 0$, $h\geq 0$.

The given condition tells us $\int_0^1 (g-h-x) dx\geq 0\Rightarrow \int_0^1 g dx\geq \int_0^1 (h+x)dx$

For the inequality we need to prove, LHS-RHS=$$\int_0^1 ((g+h)^2-x(g-h))dx\\=\int_0^1 (g^2+2gh+h^2-xg+xh)dx\\\geq \int_0^1 (g(h+x)+2gh+h^2-xg+xh)dx\\=\int_0^1 (3gh+h^2+hx)dx\geq 0$$

1

Since $\frac{1-x^2}{2} = \int_x^1 t\, dt$, then the hypothesis implies $\int_x^1 [f(t) - t]\, dt \ge 0$ for all $x\in [0,1]$. Let $F(x) := \int_x^1 [f(t) - t]\, dt$, so that $F(x) \ge 0$ for all $x\in [0,1]$. We have

$$ \int_0^1 f(x)^2\, dx - \int_0^1 2xf(x)\, dx + \int_0^1 x^2\, dx = \int_0^1 [f(x) - x]^2\, dx \ge 0$$

Thus

$$ \int_0^1 f(x)^2\, dx - \int_0^1 xf(x)\, dx \ge \int_0^1 xf(x)\, dx - \int_0^1 x^2\, dx = \int_0^1 x[f(x) - x]\, dx$$

Using integration by parts and the fact that $F(x) \ge 0$ for all $x\in [0,1]$, we find

$$\int_0^1 x[f(x) - x]\, dx = \int_0^1 x(-F'(x))\, dx = -xF(x)\bigg|_0^1 +\int_0^1 F(x)\, dx = \int_0^1 F(x)\, dx \ge 0$$

Therefore

$$\int_0^1 f(x)^2\, dx - \int_0^1 xf(x)\, dx \ge 0$$

or

$$\int_0^1 \lvert f(x)\rvert^2\, dx \ge \int_0^1 xf(x)\, dx$$

kobe
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