Are all solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

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One direction is trivial: a solution to any ODE is clearly a continuous solution to the corresponding IDE (implied differential equation).

The opposite direction seems to be false. Consider $I(f)=2\lfloor x\rfloor -1$ on the interval $[-\frac12, \frac12]$. This is solved by $f(x)=|x|$, which is continuous. But $f’=2\lfloor x\rfloor -1$ has no solutions: since RHS has a nonessential discontinuity at 0, it cannot be the derivative of any function.

The solution below doesn’t work because Darboux’s theorem assumes $f$ is differentiable. I’ve kept it for archival purposes.


When we restrict our attention to functions defined on a closed interval, we get a simple solution thanks to Darboux’s theorem. This theorem implies, in particular, that $f’$ has only essential discontinuities (this fact is mentioned on the Wiki page currently). But the fact that $f$ is a solution to an IDE on the interval implies that it has only nonessential discontinuities.

So in fact, $f’$ is continuous on the whole interval, which means that $f$ is differentiable, which means that $I(f)=D(f)$ (or, more pedantically, $I(f)(a)=\{D(f)(a)\}$ for all $a$ in the interval). This is certainly strong enough to prove the desired result 😛

I will spend some time thinking about more exotic scenarios. In particular, I have reason to believe that OP would be particularly interested in the case of half-open intervals (we talked in chat; this isn’t true). I am not quite sure how derivatives at endpoints work, so if there anyone can shore up that bit in a line or two, please mention it in the comments 🙂