Suppose we have $a_1\le a_2\le b_1\le b_2$, and $f$ has weak derivative $g_1$ on $(a_1,b_1)$ and weak derivative $g_2$ on $(a_2,b_2)$. Then by uniqueness, $g_1$ and $g_2$ must coincide on $(a_2,b_1)$. But must the resulting function (which we shall denote $g:(a_1,b_2)\to\mathbb{R}$ truly be the weak derivative of $f$?
Let $f:(a_1,b_2)\to\mathbb{R}$ be $L^1$, then can we glue weak derivatives?
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calculus
real-analysis
pde
1 Answers
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Yes, assuming $a_2 < b_1$.
Choose a test function $\varphi \in C_c^\infty(a_1,b_2)$. Then choose $\lambda_1 \in C_c^\infty(a_1,b_1)$, $\lambda_2 \in C_c^\infty(a_2,b_2)$ such that $\lambda_1+\lambda_2 = 1$ on the support of $\varphi$.
Then the definition of the weak derivative can be tested with $\lambda_1 \varphi$ on $(a_1,b_1)$ and $\lambda_2 \varphi$ on $(a_2,b_2)$ separately. Adding these two equations, you get the desired equation with $\varphi$.
On the other hand, for $a_2 = b_1$ we have to assume that the behaviour if $f$ from the left and from the right agrees. Even for $f$ smooth in $(a_1,b_1)$ and $(a_2,b_2)$, there could be a jump in $a_2 = b_1$ and then there is no weak derivative. If you want to see this case explained, ask.