Let $ f: [0,1] \to \mathbb R $ be a $ C^2 $ function such that $ f^{''} (t) \geq K $, where $ K$ is a real constant.

Can we write the above differential inequality equivalent in an integral inequality ?

$$ f(t) \leq (1- \lambda) f(0) + \lambda f(1) + \frac{K t(1-t)}{2} $$

for all $ \lambda, t \in [0,1]$.

After the replies it seems that we can't obtain such an integral bound, so my new question is if it possible to have such an equivalent way of writting the differential inequality $ f^{''} \geq K$ if we change sign to the last term, i.e.

New question: Is it true that

$$ f^{''}(t) \geq K \Longleftrightarrow f(t) \leq (1- \lambda) f(0) + \lambda f(1) - \frac{K t(1-t)}{2} \quad \forall \lambda, t \in [0,1] $$

Thank you in advance.

edit: Add a new question.

  • $\begingroup$ And what are your thoughts on the question? $\endgroup$ – Paul Sinclair Nov 3 '15 at 0:30
  • $\begingroup$ Doesn't look right: for $t = 0$ it claims that $f(0) \le f(1)$ $\endgroup$ – user58697 Nov 3 '15 at 0:31
  • $\begingroup$ @PaulSinclair: I think that I should write the points of $[0,1]$ as convex combination and integrate twice, but the problem is that i can find a way to integrate in such a way to get the desired inequality. I always have some term which causes problem. Also I want to say that is not a homowork problem. I have to use the above inequality to obtain another estimate in my problem. I just isolated this part, which is the one that I missing. Finally, I didn't came up alone with the above integral bound, I saw it and I just want to use... $\endgroup$ – passenger Nov 3 '15 at 1:25
  • $\begingroup$ ... It's a long time ago since I solve such problems and I forgot most of the tricks that we use. $\endgroup$ – passenger Nov 3 '15 at 1:25
  • $\begingroup$ @user58697: why is not right ? $\endgroup$ – passenger Nov 3 '15 at 1:29

If $f$ and $g$ are two integrable functions on $[a, b]$ with $f(x) \le g(x)$ for all $x \in [a, b]$, then $$\int_a^b f(x)\ dx \le \int_a^b g(x)\ dx.$$ In particular, if $F' = f$ and $G' = g$ and $F(a) \le G(a)$, then $F(x) = \int_a^x f(t)\ dt + F(a)$ and $G(x) = \int_a^x g(t)\ dt + G(a)$. So $F(x) \le G(x)$ everywhere in $[a, b]$. From which it follows that $$\int_a^b F(x)\ dx \le \int_a^b G(x)\ dx$$ as well.

(Edited to correct that I originally had the inequality reversed.)

To apply this to your problem: $f'' \ge K$ implies $$f'(x) - f'(0) = \int_0^x f''(t)\ dt \ge \int_0^x K\ dt = Kx$$ $$f'(x) \ge Kx + f'(0)$$ And therefore, $$f(t) - f(0) = \int_0^t f'(x)\ dx \ge \int_0^t (Kx + f'(0))dx = \frac{Kt^2}{2} + f'(0)t$$ $$f(t) \ge \frac{Kt^2}{2} + f'(0)t + f(0)$$ for all $t \in [0,1]$ (or other interval in which $f'' \ge K$).

As user58697 has pointed out, in your inequality which you hoped is true for all $t, \lambda \in [0,1]$, if we set $t = 0, \lambda = 1$, we get $f(0) \le f(1)$. This is certainly not true of every function of bounded 2nd derivative on [0,1]. If you want a specific counterexample, let $f(x) = 1 - x$.

  • $\begingroup$ Dear Paul, thank you very much for your reply ! The original inequality in my question was $ f^{''} \geq K$, so it seems that we will have the reverse inequality at the end. Can we obtain the estimate that I am asking for, if we change sign to the last term? I will edit also the question. $\endgroup$ – passenger Nov 3 '15 at 21:18
  • $\begingroup$ Yes, I see that I reversed the inequality. I'll correct that. Changing the sign of the last term of your expression is not going to save it, though. When $t = 0, \lambda = 1$, that term is $0$, so it doesn't matter what sign it has. $\endgroup$ – Paul Sinclair Nov 3 '15 at 23:17
  • $\begingroup$ I see you are right. What basically I was thinking is, if it is possible to get an inequality that "similar" to the one that I asked. When I say "similar" I mean something that looks like convex, possible with some more terms ? $\endgroup$ – passenger Nov 4 '15 at 0:19
  • $\begingroup$ It may be true if you replace $\lambda$ with $t$ in your inequality, but I haven't found the right trick to show it, or disprove it, yet. It appears to be a somewhat stronger result than I have shown above. $\endgroup$ – Paul Sinclair Nov 4 '15 at 3:42

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