Jensen's Equation Reduces to Cauchy's Equation

**Step 1: Define $g$ and check $g(0)=0$.** Let $g(x) = f(x) - f(0)$. Then $g(0) = f(0)-f(0) = 0$.

**Step 2: Rewrite Jensen's equation in terms of $g$.** Substituting $f = g + f(0)$ into $f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$ gives $g\left(\frac{x+y}{2}\right) + f(0) = \frac{g(x)+f(0)+g(y)+f(0)}{2} = \frac{g(x)+g(y)}{2} + f(0)$. The $f(0)$ terms cancel, leaving $g\left(\frac{x+y}{2}\right) = \frac{g(x)+g(y)}{2}$ — $g$ satisfies the exact same Jensen equation.

Step 3: Derive the halving identity. Set $y=0$ in the equation for $g$: $g\left(\frac{x}{2}\right) = \frac{g(x)+g(0)}{2} = \frac{g(x)}{2}$ (using $g(0)=0$ from Step 1). So $g(x/2) = g(x)/2$ for every $x$, equivalently $g(2u) = 2g(u)$ for every $u$ (substitute $u=x/2$).

**Step 4: Convert Jensen's equation for $g$ into additivity.** For arbitrary $x,y\in\mathbb{R}$, apply $g$'s Jensen equation with the pair $(x,y)$: $g\left(\frac{x+y}{2}\right) = \frac{g(x)+g(y)}{2}$. By Step 3 with $u = x+y$, the left side equals $g(x+y)/2$ (since $\frac{x+y}{2}$ is the halving of $x+y$). So $\frac{g(x+y)}{2} = \frac{g(x)+g(y)}{2}$, and multiplying both sides by $2$: $g(x+y) = g(x)+g(y)$.

Step 5: Conclude. This is exactly the additive Cauchy equation $f(x+y)=f(x)+f(y)$ applied to $g$. By Theorem 1, if $g$ (equivalently $f$, since they differ by the constant $f(0)$) is continuous, monotonic, or bounded on some interval, then $g(x) = cx$ for some constant $c=g(1)=f(1)-f(0)$. Substituting back, $f(x) = g(x) + f(0) = cx + f(0)$, the general regular solution of Jensen's equation — an affine (not necessarily linear) function.