Does fundamental theorem of calculus hold for weakly differentiable function?

That is $\int^b_a$$f'$=$f(b)-f(b)$ for $f$ being weakly differentiable??


In a sense, yes. Precisely, we have the following result:

Let $I$ be an interval (possibly unbounded) and $f\in W^{1,p}(I)$ with $1 \leq p \leq \infty$. Then there exists a function $\tilde{f}\in C(\overline{I})$ such that $f = \tilde{f}$ a.e. on $I$ and $$\int_y^x f'(t)\,dt=\tilde{f}(x)- \tilde{f}(y),\qquad \forall\ x,y\in \overline{I}.$$

Particularly, we can state:

Let $f$ be a function in $L^p(a,b)$ ($1 \leq p \leq \infty$). Assume that $f$ has a weak derivative $f'$ that also belongs to $L^p(a,b)$. Then there exists a function $\tilde{f}\in C([a,b])$ such that $f = \tilde{f}$ a.e. on $(a,b)$ and $$\int_a^b f'(t)\,dt=\tilde{f}(b)- \tilde{f}(a).$$

The proof can be found in Brezis book (Theorem 8.2).

Can you figure out why we need $\tilde{f}$ (instead of $f$) in the right hand side of the formula?


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