$f$ is uniformly continuous $\implies |f(x)| < C + D|x|$

Question

Prove $f$ is uniformly continuous $\implies$ there exist $C, D$ such that $|f(x)| < C + D|x|$.

Proof below. Please verify or critique.

By definition of uniform continuity, there exists $\delta > 0$ such that $|x_a - x_b| \leq \delta \implies |f(x_a)- f(x_b)| < 1$. Choose $D > 1/\delta$ and $C > |f(0)| + D + 1$.

For any $x$, $|f(x)| - |f(0)| \leq |f(x) - f(0)| \leq \sum_{0 < j \leq |x/\delta|+1}|f(j\delta) - f((j-1)\delta)| \leq |x/\delta|+1$, so $|f(x)| \leq |x/\delta| + 1 + |f(0)| < C + D|x|$.

Answer 1

It seems to be that you are assuming $x$ to be positive and you are trying to partition $[0, x]$ and use triangle inequalities. Be careful that $x$ can be negative and you are partitioning the interval correctly.

Let's define partition on $[0,|x|]$ such that each of the segments are at most length $\delta$.

Let $m = \left\lfloor \frac{|x|}{\delta} \right\rfloor$, $$m \le \frac{|x|}{\delta} < m+1$$

$$m \delta \le |x| < (m+1)\delta.$$

Let's divide the interval to $m+1$ segments with

$$x_i = \begin{cases} i \delta, & i \le m \\ |x|, & i = m+1\end{cases}$$

Let $s= sign(x)$,

\begin{align}|f(x)|-|f(0)| &\le |f(x)-f(0)| \\&\le \sum_{i=0}^{m}|f(sx_{i+1})-f(sx_i)|\\&\le (m+1)\\ &\le \left(\frac{|x|}{\delta}+1 \right) \end{align}

Also, I think you can just choose $C>|f(0)|+1$ though you can pick a larger number.

Answer 2

I think what you do is correct. A minor correction would be instead of $j \in \{i : 0 \le i \le \lfloor|x|/\delta\rfloor+1\}$, you could sum over $j \in \{i : 0\le i \cdot sgn(x) \le \lfloor|x|/\delta\rfloor+1\}$. This way it works for negative $x$ values as well.