1. **State the problem:** We have a function $f$ defined for all real numbers such that $$f(x + y) = f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right)$$ for all real $x, y$. We are given that $$f(8) + f(2) = 6$$ and need to find $f(128)$. 2. **Analyze the functional equation:** Substitute $y=0$ to find a simpler relation: $$f(x + 0) = f\left(\frac{x}{2}\right) + f\left(\frac{0}{2}\right) \implies f(x) = f\left(\frac{x}{2}\right) + f(0).$$ 3. Rearranging gives: $$f(x) - f\left(\frac{x}{2}\right) = f(0).$$ 4. Let $c = f(0)$. Then for any $x$, $$f(x) = f\left(\frac{x}{2}\right) + c.$$ 5. Apply this repeatedly: $$f(x) = f\left(\frac{x}{2}\right) + c = f\left(\frac{x}{4}\right) + 2c = \cdots = f\left(\frac{x}{2^n}\right) + nc.$$ 6. As $n \to \infty$, $\frac{x}{2^n} \to 0$, so $$f(x) = f(0) + nc = c + nc = c(n+1).$$ 7. But this grows without bound as $n$ increases unless $c=0$. So $c=0$ and $$f(x) = f\left(\frac{x}{2}\right).$$ 8. This means $f$ is constant for all $x$ because repeatedly halving $x$ does not change $f(x)$. Let $f(x) = k$ for some constant $k$. 9. Use the given condition: $$f(8) + f(2) = k + k = 2k = 6 \implies k = 3.$$ 10. Therefore, $$f(128) = k = 3.$$ 11. Check the options: none is 3, so re-examine the problem. The original equation is $$f(x + y) = f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right).$$ 12. Try $x = y$: $$f(2x) = 2f\left(\frac{x}{2}\right).$$ 13. Let $g(x) = f(2x)$, then $$g(x) = 2f\left(\frac{x}{2}\right) = 2f\left(\frac{x}{2}\right).$$ 14. From the original equation, $$f(x + y) = f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right).$$ 15. Substitute $x = 2a$, $y = 2b$: $$f(2a + 2b) = f(a) + f(b).$$ 16. Using $g(a+b) = f(2(a+b)) = f(2a + 2b)$, we get $$g(a+b) = f(a) + f(b).$$ 17. But $g(a+b) = f(2(a+b))$, so $$f(2(a+b)) = f(a) + f(b).$$ 18. This suggests $f$ satisfies $$f(2x + 2y) = f(x) + f(y).$$ 19. Let $h(x) = f(x)$, then $$h(2x + 2y) = h(x) + h(y).$$ 20. Set $X = 2x$, $Y = 2y$, then $$h(X + Y) = h\left(\frac{X}{2}\right) + h\left(\frac{Y}{2}\right).$$ 21. This is the original equation again, so the function is additive on scaled inputs. 22. Try a linear form $f(x) = mx$: $$f(x + y) = m(x + y)$$ $$f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right) = m\frac{x}{2} + m\frac{y}{2} = m\frac{x + y}{2}.$$ 23. Equate: $$m(x + y) = m\frac{x + y}{2} \implies m(x + y) = \frac{m}{2}(x + y).$$ 24. For all $x,y$, this implies $$m = \frac{m}{2} \implies m = 0.$$ 25. So $f(x) = 0$ is a solution, but it contradicts $f(8) + f(2) = 6$. So $f$ is not linear. 26. Try $f(x) = c$ constant: $$f(x + y) = c$$ $$f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right) = c + c = 2c.$$ 27. Equate: $$c = 2c \implies c = 0,$$ again contradicting $f(8) + f(2) = 6$. 28. Try $f(x) = kx^n$ for some $n$: $$k(x + y)^n = k\left(\frac{x}{2}\right)^n + k\left(\frac{y}{2}\right)^n = k\frac{x^n + y^n}{2^n}.$$ 29. Equate: $$(x + y)^n = \frac{x^n + y^n}{2^n}.$$ 30. Test $n=1$: $$(x + y) = \frac{x + y}{2}$$ no. 31. Test $n=0$: $$1 = \frac{1 + 1}{1} = 2$$ no. 32. Test $n=2$: $$(x + y)^2 = \frac{x^2 + y^2}{4}$$ no. 33. So no power function works. 34. Re-express original equation as $$f(x + y) = f\left(\frac{x}{2}\right) + f\left(\frac{y}{2}\right).$$ 35. Set $y = x$: $$f(2x) = 2f\left(\frac{x}{2}\right).$$ 36. Set $x = 4t$: $$f(8t) = 2f(2t).$$ 37. Set $t=1$: $$f(8) = 2f(2).$$ 38. Given $f(8) + f(2) = 6$, substitute $f(8) = 2f(2)$: $$2f(2) + f(2) = 3f(2) = 6 \implies f(2) = 2.$$ 39. Then $$f(8) = 2f(2) = 4.$$ 40. Now find $f(128)$: 41. Using $f(2x) = 2f\left(\frac{x}{2}\right)$, set $x=64$: $$f(128) = 2f(32).$$ 42. Similarly, $$f(32) = 2f(8),$$ so $$f(128) = 2 \times 2f(8) = 4f(8) = 4 \times 4 = 16.$$ 43. **Final answer:** $\boxed{16}$, which corresponds to option C.