Problem 1454

1. **Assume that $f$ satisfies (i), (ii), (iii). We will show that $\sum_{k=1}^{2008} a_k > 1$.**

From condition (iii), we have:
$$f(n) - f(n-1) = \sum_{k=1}^{2008} a_k f(n+k) - \sum_{k=0}^{2007} a_{k+1} f(n+k)$$
This can be restated as:
$$f(n) - f(n-1) = \sum_{k=1}^{2008} a_k (f(n+k) - f(n+k-1))$$

2. **Express $f(n)$ in terms of differences:**

Since $f(0) = 0$, we can write:
$$f(n) = (f(n) - f(n-1)) + (f(n-1) - f(n-2)) + \dots + (f(1) - f(0))$$
Substituting the expression from step 1, we get:
$$f(n) = a_1 (f(n+1) - f(1)) + a_2 (f(n+2) - f(2)) + \dots + a_{2008} (f(n+2008) - f(2008))$$

3. **Consider the limit as $n \to \infty$:**

Let $A = \lim_{n \to \infty} f(n)$. If we let $n$ approach infinity, the equation becomes:
$$A = A \cdot \sum_{k=1}^{2008} a_k - \sum_{k=1}^{2008} a_k f(k)$$
Since $A > 0$, we have:
$$A = A \cdot \sum_{k=1}^{2008} a_k$$
This implies:
$$\sum_{k=1}^{2008} a_k > 1$$

4. **Now assume that $\sum_{k=1}^{2008} a_k > 1$. We will show there exists a function $f$ that satisfies (i), (ii), (iii).**

Let the polynomial $g(x) = -1 + \sum_{k=1}^{2008} a_k x^k$.

5. Apply the Intermediate Value Theorem:

Note that $g(0) = -1 < 0$ and $g(1) = \sum_{k=1}^{2008} a_k - 1 > 0$. By the Intermediate Value Theorem, there exists $t \in (0, 1)$ such that $g(t) = 0$.

6. **Define the function $f(n)$:**

Let $f(n) = \sum_{k=1}^n t^k$ for all $n \in \mathbb{N}$. This already satisfies (i).

7. **Check the limit of $f(n)$ as $n \to \infty$:**

$$f(n) = \frac{t - t^{n+1}}{1 - t}$$
$$\lim_{n \to \infty} f(n) = \lim_{n \to \infty} \frac{t - t^{n+1}}{1 - t} = \frac{t}{1 - t}$$
This is a finite value since $t \in (0, 1)$. So $f$ satisfies (ii).

8. Verify condition (iii):

$$f(n) - f(n-1) = t^n$$
$$t^n = (g(t) + 1) t^n = \left( \sum_{k=1}^{2008} a_k t^k \right) t^n = \sum_{k=1}^{2008} a_k t^{n+k}$$
$$= \sum_{k=1}^{2008} a_k (f(n+k) - f(n+k-1))$$
So $f$ satisfies (iii).

Since $f$ satisfies all (i), (ii), and (iii), we are done. $\blacksquare$