3
$\begingroup$

Here's a problem I need some help with:

Let $ f $ be a twice differentiable function in the closed interval $ [-1, 1] $ and $ f''(x) \geq 0 $ for all $ x \in [-1, 1] $. Show that $$ \int^1 _{-1} f(x)dx \geq 2f(0)$$

When does the equality hold?

There was a small hint given to apply the Mean Value Theorem and the fact that $f'(x)$ is growing in that interval.

I've got a very vague idea how to apply the MVT. This is what I've done so far $$ \int_{-1} ^1 f(x) = F(1) - F(-1) = (1 -(-1))f(\xi) = 2f(\xi)$$ using the MVT. But I've got no idea how to show the inequality to be true and why $f(0)$ in particular. Actually I'm not sure if I'm on the right track to begin with.

PS. I'd prefer some hints to a complete solution at first.

$\endgroup$
  • $\begingroup$ Is it $f'(0)$ in the inequality or $f(0)$? $\endgroup$ – Alamos Apr 29 '15 at 23:02
  • $\begingroup$ $f(x)=x^2$ says it can't be $f'(0)$. $\endgroup$ – Chappers Apr 29 '15 at 23:03
  • 2
    $\begingroup$ You can consider the line $y=f'(0)x+f(0)$. The condition on the derivative tells you the graph of $f$ is above that line. Compare areas under the graphs. $\endgroup$ – Alamos Apr 29 '15 at 23:09
3
$\begingroup$

Taylor's theorem with the integral form of the remainder gives $$ f(x) = f(0) + x f'(0) + \int_0^x (x-t) f''(t) \, dt $$ (Nothing fancy here—just some integration by parts.)

The last integral is nonnegative for all $x$ by the condition on $f''$. Hence $f(x) \geqslant f(0)+xf'(0)$. Now, as Alamos notes, you integrate this inequality over $[-1,1]$.

$\endgroup$
0
$\begingroup$

Hint: Integrate by parts twice $$ \begin{align} \int_{-1}^1f(x)\,\mathrm{d}x &=\int_{-1}^0f(x)\,\mathrm{d}x+\int_0^1f(x)\,\mathrm{d}x\\ &=f(0)-\int_{-1}^0(x+1)f'(x)\,\mathrm{d}x+f(0)-\int_0^1(x-1)f'(x)\,\mathrm{d}x\\ &=2f(0)-\frac12f'(0)+\int_{-1}^0\frac{(x+1)^2}2f''(x)\,\mathrm{d}x\\ &+\frac12f'(0)+\int_0^1\frac{(x-1)^2}2f''(x)\,\mathrm{d}x\\ &=2f(0)+\int_{-1}^0\frac{(x+1)^2}2f''(x)\,\mathrm{d}x+\int_0^1\frac{(x-1)^2}2f''(x)\,\mathrm{d}x \end{align} $$

$\endgroup$

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.