Is the following inductive argument with recurrence relations valid?
Let:
- $x > 0$ be a real
- $t > 0$ be an integer
- $p_n$ be a $n$th prime
- $W_n(x)$ be the following recurrence relation:
- $W_0(x) = x$
- $W_n(x) = W_{n-1}(x) - W_{n-1}\left(\dfrac{x}{p_n}\right)$
- $H_n(x)$ be the following recurrence relation:
- $H_0(x) = \lfloor x \rfloor$
- $H_n(x) = H_{n-1}(x) - H_{n-1}\left(\dfrac{x}{p_n}\right)$
- $C_n(x)$ be the following recurrence relation:
- $C_0(x) = x - \lfloor x \rfloor = \left\{x\right\}$
- $C_n(x) = C_{n-1}(x) - C_{n-1}\left(\dfrac{x}{p_n}\right)$
Observations:
- $W_n(x) = H_n(x) + C_n(x)$ follows from:
- Base Case: $W_0(x) = x = \lfloor x \rfloor + \{x\} = H_0(x) + C_0(x)$
- Assume $W_n(x) = H_n(x) + C_n(x)$
- $W_{n+1}(x) = W_n(x) - W_n\left(\dfrac{x}{p_n}\right) = H_n(x) + C_n(x) - H_n\left(\dfrac{x}{p_n}\right) - C_n\left(\dfrac{x}{p_n}\right) = H_{n+1}(x) + C_{n+1}(x)$
- $W_n\left(\frac{x}{t}\right) = \dfrac{W_n(x)}{t}$
- Base Case: $W_0\left(\dfrac{x}{t}\right) = \dfrac{x}{t} = \dfrac{W_0(x)}{t}$
- Assume $W_n\left(\frac{x}{t}\right) = \dfrac{W_n(x)}{t}$
- $W_{n+1}\left(\dfrac{x}{t}\right) = W_n\left(\dfrac{x}{t}\right) - W_n\left(\dfrac{x}{t p_{n+1}}\right) = \dfrac{W_n(x) - W_n\left(\dfrac{x}{p_{n+1}}\right)}{t} = \dfrac{W_{n+1}(x)}{t}$
- $W_n(x) < W_{n-1}(x)$
- Base Case: $W_1(x) = W_0(x) - W_0\left(\dfrac{x}{2}\right) = x - \dfrac{x}{2} = \dfrac{x}{2} < x$
- Assume $W_n(x) < W_{n-1}(x)$
- $W_{n+1}(x) = W_n(x) - W_n\left(\dfrac{x}{p_{n+1}}\right) = \dfrac{(p_{n+1}-1)W_n(x)}{p_{n+1}} < W_n(x)$
Claim:
$$-C_n(x) < H_n(x)$$
Here's the argument:
(1) Let $u$ be a real such that $u=p_{n+1}x$
(2) $H_{n+1}(u) + C_{n+1}(u) < H_n(u) + C_n(u)$
Follows from $W_{n+1}(u) < W_n(u)$ and $W_n(u) = H_n(u) + C_n(u)$
(3) $C_{n+1}(u) - C_n(u) < H_n(u) - H_{n+1}(u)$
(4) $-C_n\left(\dfrac{u}{p_{n+1}}\right) < H_n\left(\dfrac{u}{p_{n+1}}\right)$
Since $C_{n+1}(u) = C_n(u) - C_n\left(\dfrac{u}{p_{n+1}}\right)$ and $H_{n+1}(u) = H_n(u) - H_n\left(\dfrac{u}{p_{n+1}}\right)$
(5) $-C_n(x) < H_n(x)$
Edit:
I had forgotten to call out that $x > 0$.