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Consider $\Omega = (0,2) \subseteq \mathbb{R} $ to be a support of the function $f(x)$ defined as

$$ f(x) = \begin{cases} x, &0 < x \leq 1\\ 2, &1< x \leq 2 \end{cases}$$

I need to compute the first distributional derivative of $f$.

So, I compute the following expression, $$- \int_{0}^{2} f(x) \phi ' (x) \,dx ,$$ for some test function $\phi \in C_{c}^{\infty}(\mathbb{R})$.

The above integral simplifies to

$$-\left( \int_{0}^{1} x\phi ' (x) \,dx + \int_{1}^{2} 2 \phi ' (x)\, dx \right) = \int_{0}^{1} \phi (x) \,dx + \phi(1).$$ using integration by parts.

Here this additional term $\phi(1)$ is bothering me. Can I simply ignore it considering it a constant and write the first distributional derivative $g$ as
$$ g(x) = \begin{cases} 1, &0< x\leq 1\\ 0, &\text{otherwise}. \end{cases}$$

And if this is wrong then what is the correct first distributional derivative?

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  • $\begingroup$ Do you know of a distribution who's action on a function is the same as evaluation? Perhaps that should be a term in the distributional derivative that takes the $\phi(1)$ into account. $\endgroup$ – Matt Apr 24 '17 at 14:51
  • $\begingroup$ What do you mean by "evaluation" here? $\endgroup$ – Dark_Knight Apr 24 '17 at 14:56
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    $\begingroup$ You have already written down the distributional derivative of $f$: it is the distribution $\phi \mapsto \int_0^1 \phi(x) dx + \phi(1)$. This can be written in "density form" as $1_{[0,1]}(x)+\delta(x-1)$ if you like, but this is just writing a bunch of stuff that is ultimately defined in the previously stated way anyway. $\endgroup$ – Ian Apr 24 '17 at 15:05
  • $\begingroup$ @Ian Thank you.. Got it! $\endgroup$ – Dark_Knight Apr 24 '17 at 15:32

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