# Is this an alternate way to check if a function is uniformly continuous?

My question is whether the following statement is true, or if there exists one similar.

For a differentiable real function $f:S\rightarrow \mathbb{R}$ where $S$ is an interval, and a fixed $c \in S$, let

$$g_c(x) = \int_c^x \left|\frac{df}{dt}\right|\ dt$$

Then $f$ is uniformly continuous if and only if there exists $m$ and $b$ such that $m|x|+b$ is a bound for $g_c(x)$ for all $x$

My idea was from the fact that if $f'$ is bounded, then $f$ is uniformly continuous, but the converse isn't true because of functions like $\sqrt{x}$ on $(0,\infty)$. But using the statement above, this function is strictly increasing, so $g_c(x) = \sqrt{x}-\sqrt{c}$ which is bounded by $|x|+\max(\sqrt{c}, 1)$.

The $g_c$ function came from trying to find a strictly increasing function where $g_c'(x) = |f'(x)|$. Or in general for any continuous function, I wanted to turn it into a strictly increasing function by "flipping" the function on any interval where it's decreasing, and while flipping, keeping the function continuous (sort of pivoting at local extrema), but I'm not yet sure of any way to formulate that.

• You should at least assume that the domain is an interval. Otherwise $f(x) = n + \sqrt{ x - n}$ for $x\in (n, n+1)$ is uniformly continuous, differentiable, and obviously fails your condition, with $\operatorname{dom} f = \bigcup_{n=0}^\infty (n, n+1)$. – user251257 Sep 13 '15 at 23:11
• @user251257 Thanks. That's important. – user137794 Sep 14 '15 at 0:25

Consider $f:[-1, 1]\to\mathbb R$ defined by $$f(x) = \begin{cases} x^2\cos(x^{-2}), & x\ne 0, \\ 0 , & \text{otherwise}. \end{cases}$$ Then, $f$ is continuous and differentiable with $f'(0) = 0$. However, $f'$ is not continuous at $0$, not even bounded. For $x\ne 0$ we have $$f'(x) = 2x\cos(x^{-2}) + 2x^{-1}\sin(x^{-2}).$$ For $-1 \le c \le 0 < x \le 1$, we obtain \begin{align} \int_c^x |f'(t)| \; \mathrm dt &\ge \int_{c}^x |2t^{-1} \sin(t^{-2})| \; \mathrm d t - \underbrace{\int_c^x |2t\cos(t^{-2})|\;\mathrm dt}_{\le C < \infty} \\ &\ge \int_{0}^x |2t^{-1} \sin(t^{-2})| \; \mathrm d t - C \\ &\ge \int_{x^{-1}}^{\infty} \left| \frac{\sin(t^2)}{t} \right| \; \mathrm d t - C = \infty, \end{align} where $$C = \int_{-1}^1 |2t\cos(t^{-2})| \;\mathrm dt.$$ The cases $-1 \le c < 0 \le x \le 1$ and $-1 \le x < c \le 1$ are similar.
Now, if you like, we could extend $f$ to a uniform continuous and differentiable function on $\mathbb R$ by some affine linear functions on both side.
• Should that be $f' = 2\cos(x^2) - [\sin(x^2)/x^2]$? – user137794 Sep 14 '15 at 1:06
• I don't think is function is uniformly continuous to begin with. At the end points, it tends to $x^2$ which isn't uniformly continuous. – user137794 Sep 14 '15 at 1:58
• @user137794: $f$ is a continuous functions on a compact interval... – user251257 Sep 14 '15 at 1:59