Which of the following are uniformly continuous over $\mathbb{R}$?

Question

Which of the following are uniformly continuous over $\mathbb{R}$?

(i) $f(x)=\int_{0}^{x}g(t)dt$, where $g:\mathbb{R}\to\mathbb{R}$ is a continuously differentiable function.

(ii) $f(x)=\int_{0}^{x}g(t)dt$, where $g:\mathbb{R}\to\mathbb{R}$ is a continuous function which is bounded.

My attempt:

(ii) $|\int_{0}^{x}g(t)dt- \int_{0}^{y}g(t)dt|=|g(c)||x-y|$ for some $c<x<y$, this follows from MVT. Since $g(x)$ is a continuous fn which is bounded, this implies that $|\int_{0}^{x}g(t)dt- \int_{0}^{y}g(t)dt|\le M|x-y|$. Hence, uniformly continuous.

I can have the same approach to (i), but we don't know whether $|g(c)|$ is bounded or not, in that case. Please help me in solving the (i) part, whether it is uniformly continuous over $\mathbb{R}$. I believe that the justification for (ii) is correct. Thank you.

The answer is: only (ii) is uniformly continuous.

Answer 1

your answer seems correct for part (ii).For part (i)Take $g(t) = t$ then your $ f(x)$ will be $\frac{x^2}{2}$ which is not a uniformly continuous function over $\Bbb R$

Answer 2

$|\int_0^x g(t) dt-\int_0^y g(t) dt|=|\int_y^xg(t)dt|=|g(c)||x-y|$ for some c between x and y non inclusive. The second equality is true because of MVT. Note that g(x) and g(y) are assumed to have finite value, and therefore g(c) is finite too. Thus we have proven $|f(x)-f(y)|$ is bounded.

Answer 3

For (ii), the derivative $f'(x)=g(x)$ which is bounded on $\mathbb R$ so $f(x)$ is uniformly continuous on $\mathbb R$.

To discard (i), take $g(t)=e^t$.