Let $f$ be a function that has a weak derivative on $(a,b)$ and assume that there is $c$ with $a < c < b$ such that it satisfies $f\bigg|_{(a,c]} \in C^1(a,c]$ and $f\bigg|_{(a,c]} \in C^1[c,b)$, how do I show that in this case we must have $f \in C^0(a,b)$? In other words, $f$ is continuous.
My thoughts on this, this seems almost immediate, if $f$ restricted to $(a,c]$ and $[c,b)$ is already continuous, then we only need to worry about it at the point $x=c$. Is this correct reasoning? If so, how does the weak derivative come into play?
Also, is it true if $f$ satisfies these conditions, i.e., $f\bigg|_{(a,c]} \in C^1(a,c]$ and $f\bigg|_{(a,c]} \in C^1[c,b)$, then $f$ must admit a weak derivative? It seems we would have to construct the function $g$ so that $$\int f(t) \phi'(t) \, dt = - \int g(t) \phi(t) \, dt.$$ How does one usually go about finding weak derivatives or I suppose in this case constructing them.
If you treat $f$ as a function $(a,b) \longrightarrow \mathbb R$, saying $\left. f \right|_{(a,c]} \in C^1$ and $\left. f \right|_{[c,b)} \in C^1$ you automatically have $$ f \in C^0(a,c] \cap C^0[c,b) = C^0(a,b). $$ If you treat $f$ as a $L^1_{loc}(a,b)$ object, then you only have that $f$ has a $C^1$ representative on $(a,c]$ and another $C^1$ representative on $[c,b)$. This does not yield continuity in general, as the example $\chi_{(a,c]}$ shows.
I guess that what you are referring to "$f$ has a weak derivative" stands for $f \in W^{1,1}(a,b)$, possibly $f \in W^{1,1}_{loc}(a,b)$. In any case you may know that $$ W^{1,1}_{loc}(a,b) \subset C^0(a,b) $$ so that the question you are addressing is fully understood.