Intuitively, I understand that this is not possible, but I'm trying to provide a concise mathematical explanation.
$f(x) > 0$: This means the function is always positive.
$f'(x) < 0$: This means the function is always decreasing.
$f''(x) < 0$: This means the function is always concave down.
If the function is always decreasing and concave down, then at some point it will cross the x axis which would violate the property $f(x) > 0$.
Would this be enough to consider this problem solved?
Would this be enough to consider this problem solved?
To prove that such a $f$ exists, the simplest way to do so is exhibit a specific example $f$ (like $f(x) = 2x+1$ or $f(x) = \log{x}$ or whatever) and prove the inequalities hold.
To prove that no such $f$ exists, one way to do so is to start by saying:
and then derive a contradiction somehow.
Your intuition is correct, but not nearly formalized enough to constitute proof. Here's how I might go about formally proving your statement using that key insight your intuition gave you about the derivative decreasing:
From our condition, we have that $f'(0) < 0.$ Because $f''(x) < 0$ for all $x,$ for $x > 0$ we must have that $f'(x) < f'(0)$: intuitively we know this because $f'$ must be decreasing, and more formally we can say that if $f'(x_0) \geq f'(0)$ for any $x_0 > 0,$ then the Mean Value Theorem would indicate that there must exist some $c$ between $0$ and $x_0$ such that $f''(c) = \frac{f'(x_0) - f'(0)}{x_0} \geq 0,$ which contradicts our condition that $f''(x) < 0.$
So, if $f'(x) < f'(0)$ for $x > 0,$ then we must have that $\int_0^x f'(t) dt < \int_0^x f'(0) dt$ as well for $x > 0$, so $f(x) - f(0) < f'(0)x,$ implying $f(x) < f'(0)x + f(0).$ This is a typical consequence of being concave down.
It's clear that $x > -\frac{f(0)}{f'(0)} > 0$ will cause $f'(0)x + f(0) < 0,$ so we must have $f(x) < 0$ there as well by transitivity.
This would conclude the proof that the properties $f'(x) < 0$ and $f''(x) < 0$ imply that $f$ must go below $0$ at some point. If you have any questions about the steps though, feel free to let me know: in particular I know I skipped a bit of the algebra at the end but I think you should be able to fill in the blanks.
To make it rigorous, one can use a line to $y = f(x)$.
Let $a>0$ and take a line tangent to $y=f(x)$ at $(a, f(a)$. Namely, the line can be expressed as $y = f^\prime(a)(x-a) + f(a)$. We now show $y = f(x)$ is below the tangent line for $x>a$.
Let $g(x) = f(x) - \{f^\prime(a)(x-a) + f(a)\}$. Then $$ g^\prime(x) = f^\prime(x) - f^\prime(a), g^{\prime\prime}(x) = f^{\prime\prime}(x)<0 $$
Since $g^{\prime\prime}(x) <0$, we must have $g^\prime(x) < g^\prime(a) = f^\prime(a) - f^\prime(a)=0$ for $x >a$. Thus we have shown $g(x)<g(a) = f(a) - \{f^\prime(a)(a-a) + f(a)\} = 0$ for $x<a$.
This is equivalent to $f(x) < f^\prime(a)(x-a) + f(a)$ for $x>a$.
Since $f^\prime(a)<0$, $f^\prime(a)(x-a) + f(a)<0$ for $x > a - f(a)/f^\prime(a)$.
This means $f(x) < 0$ for $x > a - f(a)/f^\prime(a)$.
Let $f(0) = b \gt 0$ and $f^\prime(0) = a \lt 0$. As $f$ is concave, it is below any of its tangents. Therefore $f(x) \lt ax + b$ for $x \gt 0$. In particular $f(x) \lt 0$ for $x \gt \frac{-b}{a}$. A contradiction with the first hypothesis.
Assume such an $f$ exists. Let $g(x) = f(x) - f(0)$, then $g$ is strictly concave, and $$ 0 \leq g(x)<g(0) + g'(0)x=g'(0)x<0 $$ for any $x > 0$, a contradiction.
