Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that:
$\bullet$ $f(x)<2$ for all $x\in (0,1)$;
$\bullet$ for all real numbers $x,y$ we have:
$$max\{f(x+y),f(x-y)\}=f(x)+f(y)$$

Proposed by Navid Safaei
**
   The proposed solution is \( f(x) = c|x| \) for all \( x \in \mathbb{R} \) and \( c \in [0,2] \). We need to verify that this function satisfies both conditions.

   - **Condition 1:**
     For \( x \in (0,1) \), we have \( f(x) = c|x| = cx \). Since \( c \in [0,2] \), it follows that \( cx < 2 \). Thus, \( f(x) < 2 \) for all \( x \in (0,1) \).

   - **Condition 2:**
     We need to check that \( \max \{ f(x + y), f(x - y) \} = f(x) + f(y) \) holds for all \( x, y \in \mathbb{R} \).

     - If \( x \) and \( y \) have the same sign:
       \[
       \max \{ f(x+y), f(x-y) \} = \max \{ c|x+y|, c|x-y| \}
       \]
       Since \( x \) and \( y \) have the same sign, \( |x+y| \geq |x-y| \). Therefore,
       \[
       \max \{ c|x+y|, c|x-y| \} = c|x+y| = c(|x| + |y|) = c|x| + c|y| = f(x) + f(y)
       \]

     - If \( x \) and \( y \) have opposite signs:
       \[
       \max \{ f(x+y), f(x-y) \} = \max \{ c|x+y|, c|x-y| \}
       \]
       Since \( x \) and \( y \) have opposite signs, \( |x-y| \geq |x+y| \). Therefore,
       \[
       \max \{ c|x+y|, c|x-y| \} = c|x-y| = c(|x| + |y|) = c|x| + c|y| = f(x) + f(y)
       \]

3. **Proof of Uniqueness:**
   We need to show that \( f(x) = c|x| \) for \( c \in [0,2] \) is the only solution.

   - **Step 1: Basic Properties**
     Let \( P(x, y) \) denote the assertion of the functional equation:
     \[
     \max \{ f(x+y), f(x-y) \} = f(x) + f(y)
     \]
     - \( P(0,0) \) gives \( \max \{ f(0), f(0) \} = 2f(0) \), so \( f(0) = 0 \).
     - \( P(0, x) \) gives \( \max \{ f(x), f(-x) \} = f(x) \), so \( f(x) \geq f(-x) \). Similarly, \( P(0, -x) \) gives \( f(-x) \geq f(x) \). Thus, \( f(x) = f(-x) \) for all \( x \in \mathbb{R} \).

   - **Step 2: Homogeneity**
     We claim that \( f(nx) = |n|f(x) \) for any \( n \in \mathbb{Z} \) and \( x \in \mathbb{R} \).

     - For \( n = 1 \), it is trivial.
     - For \( n = 2 \), \( P(x, x) \) gives \( \max \{ f(2x), 0 \} = 2f(x) \), so \( f(2x) = 2f(x) \).
     - By induction, assume \( f(kx) = kf(x) \) for some \( k \in \mathbb{N} \). Then \( P(kx, x) \) gives:
       \[
       \max \{ f((k+1)x), f((k-1)x) \} = f(kx) + f(x) = (k+1)f(x)
       \]
       Since \( f \) is non-negative, \( f((k+1)x) = (k+1)f(x) \).

     - For negative integers, use \( f(-x) = f(x) \).

   - **Step 3: Rational Arguments**
     We extend the result to rational arguments. For \( m = \frac{p}{q} \) with \( p, q \in \mathbb{Z} \), we have:
     \[
     q f \left( \frac{p}{q} x \right) = f(px) = pf(x) \implies f \left( \frac{p}{q} x \right) = \frac{p}{q} f(x)
     \]

   - **Step 4: Continuity and Linearity**
     Since \( f \) is additive and homogeneous on rationals, and \( f(x) \geq 0 \) for \( x \geq 0 \), it follows that \( f \) is linear on \( \mathbb{R}^+ \). Thus, \( f(x) = cx \) for \( x \geq 0 \) and some \( c \geq 0 \).

     Given \( f(x) < 2 \) for \( x \in (0,1) \), we have \( c \leq 2 \). Therefore, \( f(x) = c|x| \) for \( c \in [0,2] \).

Conclusion:
\[
\boxed{f(x) = c|x| \text{ for } c \in [0,2]}
\]