MHB Can you prove Tate's theorem on bounded mappings?

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Here is this week's POTW:

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Prove the following theorem of Tate: If $\phi : X\to Y$ is a mapping of real Banach spaces such that for some positive number $M$, $\lvert \phi(x + y) - \phi(x) - \phi(y)\rvert \le M$ ($x,y\in X$), then there is a unique additive mapping $\psi : X \to Y$ such that $\psi - \phi$ is bounded on $X$.

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No one answered this week's problem. You can read my solution below.

Spoiler

Note $\phi(2x) = 2\phi(x) + O(x)$, where $\lvert O(x) \lvert < M$. Inductively, $\phi(2^nx) = 2^n\phi(x) + (2^n-1)O(x)$ for all positive integers $n$. Fix $x\in X$ and consider the sequence $\phi_n := \frac{\phi(2^nx)}{2^n}$, $n\in \Bbb N$. Since

$$|\phi_n - \phi_m| = \left\lvert \left[\phi(x) + \left(1 - \frac{1}{2^n}\right)O(x)\right] - \left[\phi(x) + \left(1 - \frac{1}{2^m}\right)O(x)\right] \right\rvert = \left\lvert \left(\frac{1}{2^m} - \frac{1}{2^n}\right)O(x)\right\rvert = \left(\frac{1}{2^m} - \frac{1}{2^n}\right)O(x) \to 0$$

as $n,m\to \infty$, then $\phi_n$ is a Cauchy sequence in $Y$. Since $Y$ is complete and $x$ was arbitrary, there is a mapping $\psi : X\to Y$ such that $\lim\limits_{n\to \infty} \phi_n(x) = \psi(x)$ for all $x\in X$. Now

$$\psi(x + y) = \lim_{n\to \infty} \frac{\phi(2^nx + 2^n y)}{2^n} = \lim_{n\to \infty} \frac{\phi(2^nx) + \phi(2^ny) + E(x, y)}{2^n} = \psi(x) + \psi(y) + \lim_{n\to \infty} \frac{E(x,y)}{2^n},$$

where $\lvert E(x,y)\rvert < M$. Thus $\lim\limits_{n\to \infty} \frac{E(x,y)}{2^n} = 0$ and $\psi$ is additive.

To see that $\psi - \phi$ is bounded, fix $x\in X$, and choose a positive integer $N = N(x)$ such that $\lvert \psi(x) - \phi_N(x)\rvert < M$. Then

$$\lvert \psi(x) - \phi(x)\rvert \le \lvert \psi(x) - \phi_N(x)\rvert + \lvert \phi_N(x) - \phi(x)\rvert < M + M = 2M$$

Since $x$ was arbitrary, $\psi - \phi$ is bounded on $X$.

For uniqueness, let $\theta$ be any additive mapping from $X$ to $Y$ such that $\theta - \phi$ is bounded on $X$. Given $x\in X$, $\theta(x) = \frac{\theta(2^n x)}{2^n}$ for all $n\in \Bbb N$. Thus

$$\lvert \theta(x) - \phi_n(x)\rvert = \left\lvert \frac{\theta(2^nx) - \phi(2^nx)}{2^n}\right\rvert \le \frac{C}{2^n},$$

where $C$ is a constant. Hence

$$\theta(x) = \lim_{n\to \infty} \phi_n(x) = \psi(x)$$