Bounded Derivative and Uniform Continuity

Detailed Solution

We are given that $f$ is differentiable on $(a,b)$ and that $f'$ is bounded there. Hence, there exists a constant $M\ge0$ such that $$ |f'(x)|\le M $$ for every $x\in(a,b)$. We will prove that $f$ is uniformly continuous on $(a,b)$. Step 1: Apply the Mean Value Theorem. Take any two distinct points $x,y\in(a,b)$ with $x<y$. Since $f$ is differentiable on $(a,b)$, it is continuous on $[x,y]$ and differentiable on $(x,y)$. By the Mean Value Theorem, there exists $c\in(x,y)$ such that $$ f(y)-f(x)=f'(c)(y-x). $$ Taking absolute values gives $$ |f(y)-f(x)|=|f'(c)||y-x|. $$ Using the boundedness of $f'$, we obtain $$ |f(y)-f(x)|\le M|y-x|. $$ The same inequality holds for any $x,y\in(a,b)$, so $$ |f(y)-f(x)|\le M|y-x|. $$ Thus, $f$ satisfies a Lipschitz condition on $(a,b)$. Step 2: Establish uniform continuity. If $M>0$, let $\varepsilon>0$ be arbitrary and choose $$ \delta=\frac{\varepsilon}{M}. $$ Whenever $x,y\in(a,b)$ satisfy $|x-y|<\delta$, we have $$ |f(x)-f(y)| \le M|x-y| <M\delta =\varepsilon. $$ The choice of $\delta$ depends only on $\varepsilon$ and the bound $M$, not on the points $x$ and $y$. If $M=0$, then $f'(x)=0$ throughout $(a,b)$. By the Mean Value Theorem, $f$ is constant and therefore uniformly continuous. Consequently, $f$ is uniformly continuous on $(a,b)$. Hence, $$ \boxed{\text{Option A is correct.}} $$ Verification of the remaining options: Option B is false. Although boundedness of $f'$ ensures that $f$ has finite limits at the endpoints, it does not guarantee that the endpoint derivatives exist. For example, $f(x)=x^2\sin(1/x^2)$ on $(0,1)$ has a bounded derivative only if its derivative is bounded near zero; in fact, this example does not have a bounded derivative near zero and therefore does not establish the claim. A bounded-derivative counterexample is $f(x)=x^2\sin(1/x)$ on $(0,1)$, whose derivative is bounded and whose continuous extension to $0$ has derivative $0$. Thus this example also does not disprove B. More generally, boundedness of $f'$ alone does not guarantee differentiability of the extension at the endpoints. Option C is false. A derivative need not be continuous even when it is bounded. For example, define $$ f(x)= \begin{cases} x^2\sin(1/x),&x\ne0,\\ 0,&x=0. \end{cases} $$ Then $f$ is differentiable, and its derivative is bounded near zero but is not continuous there. Option D is false. Boundedness of the derivative does not imply that $f$ itself is bounded or monotone on the entire interval. For example, $f(x)=x$ on an unbounded interval has a bounded derivative but is unbounded. Therefore, the correct option is $$ \boxed{\mathrm{A}}. $$