Primitives on $[a,c)$ and $(c,b]$ implies primitives on $[a,b]$

Question

Let $a<c<b \in \mathbb R$ and $f:[a,b] \to \mathbb R$ be continuous in c. If $f$ has primitives on $[a,c)$ and $(c,b]$ then $f$ has primitives on $[a,b]$. Can somebody help me, please? I have no idea how to start the proof.

Answer 1

Well, if f is continuous, it always has a primitive on [a,b]. For instance $F(x) = \int_a^x f(t)dt$ is one such primitive.

Edit : Misread your question. Let's call $F_1$, and $F_2$ the two primitives of f. Since both are differentiable, they are both continuous. In particular, they are continuous in c. WLOG we can suppose $F_1(c)=F_2(c)$. Then, let's consider $F(x)=F_1(x)$ if x < c and $F_2(x)$ otherwise. We will check that F is a primitive of f. We just need to compute the derivative of F at c, and check that it is indeed $f(c)$. By our hypothesis our left derivative at c is the left derivative of $F_1$ at c which is equal to f(c) by continuity of f at c. Same argument for right derivative shows that it's also equal to f(c). As a result, F is differentiable at c and F'(c) = f(c), hence F is a primitive of f on [a, b]

Answer 2

A simple application of Lagrange is the fact that if $g$ is continuous at $c$, differentiable in $(a,c)\cup (c,b)$ and $\lim_{x\to c}g'(x)=L$, then $g$ is differentiable on $(a,b)$ and $g'(c)=L$.

Therefore, we just need to prove that there is some primitive $F$ of $f$ on $[a,c)\cup(c,b]$ which extends to a continuous function $\overline F:[a,b]\to\Bbb R$.

Since, given some two functions $F_-:[a,c)\to\Bbb R$ and $F_+:[a,c)\to\Bbb R$ such that $F_-'=f$ on $[a,c)$ and $F_+'=f$ on $(c,b]$, a function $F$ is a primitive of $f$ on $[a,c)\cup (c,b]$ if and only if there are two constants $L_1, L_2$ such that $$F(x)=\begin{cases}F_-(x)+L_1&\text{if }x\in [a,c)\\ F_+(x)+L_2&\text{if }x\in (c,b]\end{cases},$$ we can (why?) reduce the discussion to deciding whether some primitive $F_+$ of $f$ on $(c,b]$ extends to a continuous function $\overline F_+$ on $[c,b]$.

Now, a useful lemma that comes to mind is the fact that a uniformly continuous function $g:X\to \Bbb R$, where $X$ some subset of $\Bbb R$ (or a metric space $M$) extends to a continuous function $\overline g:\overline X\to\Bbb R$.

Uur $F_+$ may not be immediately recognizable as uniformly continuous on $(c,b]$, but since $F_+'=f$ is bounded in some right-neighbourhood $(c,c+\varepsilon)$ of $c$ by hypothesis, $F_+'$ is certainly Lipschitz continuous on $(c,c+\varepsilon)$, and therefore it extends continuously to $[c,c+\varepsilon]$ (and thus to $[c,b]$).

The ingredients are these. The details must be filled in.