6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your last one but step is incorrect because the inequality for the second term is going the other way. $\endgroup$ – Aravind Sep 29 '15 at 6:21
  • $\begingroup$ 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? $\endgroup$ – Dair Sep 29 '15 at 6:24
  • $\begingroup$ 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 $x$. $\endgroup$ – Aravind Sep 29 '15 at 6:41
  • $\begingroup$ You should probably apply Cauchy-Schwarz to the pair of functions $f(x)$ and $x$. $\endgroup$ – Aravind Sep 29 '15 at 6:56
  • $\begingroup$ 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: $$\int_0^1 |f(x)|^2 dx \geq \int_0^1 x f(x) dx$$ $\endgroup$ – Dair Sep 29 '15 at 8:37
1
$\begingroup$

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=$$\int_0^1 ((g+h)^2-x(g-h))dx\\=\int_0^1 (g^2+2gh+h^2-xg+xh)dx\\\geq \int_0^1 (g(h+x)+2gh+h^2-xg+xh)dx\\=\int_0^1 (3gh+h^2+hx)dx\geq 0$$

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.