Let $c$ be a real number. If the inequality
$$f(c)\cdot f(-c)\ge f(a)$$
holds for all $f(x)=x^2-2ax+b$ where $a$ and $b$ are arbitrary real numbers, find all possible values of $c$.
1. **Define the function and inequality:**
   Given the function \( f(x) = x^2 - 2ax + b \), we need to find all possible values of \( c \) such that the inequality \( f(c) \cdot f(-c) \ge f(a) \) holds for all real numbers \( a \) and \( b \).

2. **Evaluate \( f(c) \) and \( f(-c) \):**
   \[
   f(c) = c^2 - 2ac + b
   \]
   \[
   f(-c) = c^2 + 2ac + b
   \]

3. **Form the product \( f(c) \cdot f(-c) \):**
   \[
   f(c) \cdot f(-c) = (c^2 - 2ac + b)(c^2 + 2ac + b)
   \]
   Using the identity \((x-y)(x+y) = x^2 - y^2\), we get:
   \[
   f(c) \cdot f(-c) = (c^2 + b)^2 - (2ac)^2 = c^4 + 2bc^2 + b^2 - 4a^2c^2
   \]

4. **Set up the inequality:**
   \[
   c^4 + 2bc^2 + b^2 - 4a^2c^2 \ge a^2
   \]
   Rearrange to form a quadratic in \( b \):
   \[
   b^2 + (2c^2 - 1)b + c^4 - 4a^2c^2 + a^2 \ge 0
   \]
   Let \( g(b) = b^2 + (2c^2 - 1)b + c^4 - 4a^2c^2 + a^2 \).

5. **Analyze the quadratic \( g(b) \):**
   For \( g(b) \ge 0 \) to hold for all \( b \), the discriminant of \( g(b) \) must be non-positive:
   \[
   \Delta = (2c^2 - 1)^2 - 4(c^4 - 4a^2c^2 + a^2)
   \]
   Simplify the discriminant:
   \[
   \Delta = (2c^2 - 1)^2 - 4(c^4 - 4a^2c^2 + a^2)
   \]
   \[
   \Delta = 4c^4 - 4c^2 + 1 - 4c^4 + 16a^2c^2 - 4a^2
   \]
   \[
   \Delta = -4c^2 + 1 + 16a^2c^2 - 4a^2
   \]
   \[
   \Delta = (4a^2 - 1)(4c^2 - 1)
   \]

6. **Determine the conditions for the discriminant:**
   For the discriminant to be non-positive:
   \[
   (4a^2 - 1)(4c^2 - 1) \le 0
   \]
   This inequality holds if and only if \( 4c^2 - 1 = 0 \), because otherwise, we can choose \( a \) such that \( \text{sgn}(4a^2 - 1) = \text{sgn}(4c^2 - 1) \), making the discriminant positive.

7. **Solve for \( c \):**
   \[
   4c^2 - 1 = 0
   \]
   \[
   4c^2 = 1
   \]
   \[
   c^2 = \frac{1}{4}
   \]
   \[
   c = \pm \frac{1}{2}
   \]

The final answer is \( \boxed{ c = \pm \frac{1}{2} } \)