Does every continuous function has an anti-derivative?

Solution 1:

It is not true that every continuous function has an antiderivative. Its domain matters. No continuous function $f:[a,b]\to\mathbb R$ has an antiderivative, since such an antiderivative would have to be differentiable on a non-open set, which is impossible by definition. The derivative is only defined at interior points of the domain.

However, if we allow one-sided differentiability, which some authors do, then continuous functions on closed intervals do have an antiderivative, given by the integral function $I_{f,a}$, which integrates starting at $a$. This is differentiable (from the right) at $a$, since

$$\begin{align} \frac{I_{f,a}(x)-I_{f,a}(a)}{x-a}&=\frac{1}{x-a}\left(\int_a^x f(t)\mathrm dt-\int_a^{a}f(t)\mathrm dt\right)\\ &=\frac{1}{x-a}\int_a^x f(t)\mathrm dt\\ &=f(a)-f(a)+\frac{1}{x-a}\int_a^x f(t)\mathrm dt\\ &=f(a)+\frac{1}{x-a}\int_a^x f(t)-f(a)\mathrm dt. \end{align}$$

The absolute value of the term with the integral in the end is bounded by $\frac{\vert x-a\vert\sup_{t\in[a,x]}\vert f(t)-f(a)\vert}{\vert x-a\vert}=\sup_t\vert f(t)-f(a)\vert$, which, by continuity of $f$, goes to $0$ as $x\to a$, and then the difference quotient above goes to $f(a)$. By a similar argument it is differentiable (from the left) at $b$.

Note that this is essentially the proof of the FTC, with no alterations.

If you don't want to go the route of one-sided differentiability, then the theorems like integration by parts need more careful consideration of the boundary points of the domain. You would need to show that the functions appearing in the theorems can be continuously extended to the boundaries, and then by uniqueness of the continuous extension you could claim that the equations hold for those extensions, too. For instance, $$\int_\alpha^\beta f'(t)g(t)\mathrm dt=f(\beta)g(\beta)-f(\alpha)g(\beta)-\int_\alpha^\beta f(t)g'(t)\mathrm dt$$ holds for all $a<\alpha<\beta<b$, and if the two sides of the equation as functions of $\alpha$ and $\beta$ can be continuously extended to $(a,b)$ (to be read as a tuple, not an interval), then the equations also hold for the continuous extensions, which we write by replacing the Greek letters by their Latin counterparts.