This answer actually answers your question, too. (See also point (2) here.)
Yes. The antiderivative of an integrable function is absolutely continuous. If $f$ is $C^1$ and of bounded variation, then $\int \lvert f'\rvert = V(f) < \infty$. So $f$ is the antiderivative of an integrable function.
It is sufficient to prove that $f$ is constant in $]-\infty,+\infty[$.
If it was not constant there would exists $x_0\in \mathbb{R}$ such that $f’(x_0)\ne0 $.
There are only two possibilities: $f’(x_0)>0$ or $f’(x_0)<0$.
Without loss of generality we can suppose that $f’(x_0)>0$.
The hypothesis $f’’(x)\ge0$ for all $x\in \mathbb{R}$ implies that $f’(x)$ is a monotone nondecreasing function, so $f’(x)\ge f’(x_0)>0$ for all $x\in ]x_0,+\infty[$.
Moreover for all $x \in ]x_0, +\infty[$ we can apply Langrange Theorem to the interval $[x_0,x]$, so there exists $c \in ]x_0,x[$ such that $f(x)-f(x_0)=f’(c)(x-x_0)\ge f’(x_0)(x-x_0)$.
It means that $f(x)\ge f(x_0)+f’(x_0)(x-x_0)$ for all $x \in ]x_0, +\infty[$.
So $f$ is not bounded from above in $]x_0, +\infty[$, but it is absurd because it contradicts one of the hypothesis.
Hence it is impossible that $f$ is not constant in $]-\infty,+\infty[$.
It means $f$ is constant in $]-\infty,+\infty[$, so it is infinitely differentiable.
Best Answer
No, e.g. $\sin (x^2){}{}{}{}$.