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Let $f$ be a function of class $C^2$ from $ \mathbb{R^{+,*}}$ to $\mathbb{R^{+, *}}$ (where $\mathbb{R^{+,*}}$ denotes the set of nonnegative reals) such that $f''>f$ and $f$ is bounded above. Show that $f \le f(0)e^{-x}$.

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    The closest I see among the "big results" is Gronwall's lemma (https://www.encyclopediaofmath.org/index.php/Gronwall_lemma). – 2017-01-09
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    With $h=f''-f$, there exists two constants $a$ and $b$ such that $$\forall x, f(x) = \left( a -\frac 12 \int_0^x h(t)e^t dt \right) e^{-x} + \left( b +\frac 12 \int_0^x h(t)e^{-t} dt\right) e^{x}$$ I can't make much progress unless something is known about $h$ (boundedness would help)... – 2017-01-09
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    @LeGrandDODOM at very first glance, doesn't the fact that $f$ is bounded imply $b=0$? – 2017-01-09
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    @ClementC.: Indeed,... and since $f(0)=a+b=a$, the problem comes down to showig $(a-y)e^{-x}+ye^x \le ae^{-x}$ (where $y$ is the expression in the intregral multiplied by $1/2$) equivalent to $e^{2x} \le 1$, which is false for $x>0$. There is a problem somewhere... – 2017-01-09
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    On the otherhand, how did you arrive at that expression for $f$ in terms of $h$. The homogeneous solution component is correct but it seems to me that the specific solution component should be a double integral, since the equation is of degree $2$. And assuming that expression is correct we actually arrive at the opposite of what the problem wants us to show. – 2017-01-09
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    @JoshuaBenabou I used parameter variation method on the equation $f''-f=h$ where $h$ is an arbitrary function. I checked my computations again and did not find any mistake. How do you prove that $b$ must be $0$ ? Also you can't write $(a-y)e^{-x}+ye^x \le ae^{-x}$. The integrals are not the same, one has $e^t$ in it whereas the other has $e^{-t}$ in it. – 2017-01-09
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    Oh dear. I read quickly and thought they were the same. The deduction that $b=0$ is also false... – 2017-01-09
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    @JoshuaBenabou Where did you find this problem? It seems very challenging – 2017-01-09
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    Apparently its an ancient problem from French prepa, probably Oral X. – 2017-01-09

2 Answers 2

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$f'$ is increasing because $f''(x) > f(x) \ge 0$. Then $f'$ must be negative because $f$ would be unbounded otherwise.

First define $g(x) = e^{-x} (f'(x) + f(x))$. Then $$ g'(x) = e^{-x}(f''(x) - f(x)) > 0 $$ so that $g$ is increasing. It follows that for $ x < a$ $$ e^{-x} (f'(x) + f(x)) \le e^{-a} (f'(a) + f(a)) \le e^{-a}M $$ where $M$ is an upper bound for $f$. Letting $a \to \infty$ we get that the left-hand side is $\le 0$, i.e. $$ f'(x) + f(x) \le 0 \text{ for all } x \ge 0 \, . $$

Finally define $h(x) = e^x f(x)$. Then $$ h'(x) = e^x(f'(x) + f(x)) \le 0 $$ so that $h$ is decreasing. It follows that $$ e^x f(x) = h(x) \le h(0) = f(0) \, . $$

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    In the last line, you used $f'(0)\geq f'(x)$, but you said at the beginning that $f'$ is increasing ... – 2017-01-09
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    @LeGrandDODOM: You are completely right, that was a silly error. – I have given it another try, hopefully it is correct now. (The underlying idea is probably the same as in the other answer.) – 2017-01-09
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Let $0\le a \le x$. It follows that $$ f''(x)-f'(x)>-f'(x)+f(x) \\ \frac{d}{dx}(e^{-x}f'(x))>-\frac{d}{dx}(e^{-x}f(x)) \\ e^{-x}f'(x)-e^{-a}f'(a)>-e^{-x}f(x)+e^{-a}f(a) \\ \frac{d}{dx}(e^{x}f(x))>e^{2x-a}(f'(a)+f(a)) \\ e^{x}f(x)-e^{a}f(a)>\frac12e^{-a}(e^{2x}-1)(f'(a)+f(a)) \\ f(x)>e^{a-x}f(a)+\frac12(e^{x-a}-e^{-x-a})(f'(a)+f(a)) $$ As the left side is bounded, so is the right side. This can only be if $$ \frac12e^{x-a}(f'(a)+f(a))\le0, $$ and this must hold for all $a>0$.

Thus $\frac{d}{dx}(e^xf(x))\le0$ which directly leads to the claim.

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    Unless I'm mistaken, $f'(x) – 2017-01-09
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    @LeGrandDODOM : I was of the wrong opinion that $\frac{d}{dx}(e^{2x}\frac{d}{dx}(e^{-x}f(x)))>0$ and $\frac{d}{dx}(e^{-2x}\frac{d}{dx}(e^{x}f(x)))>0$ integrate to equivalent bounds. Now it is the other way around and this time the necessary condition really leads to the claim. – 2017-01-09