What is a derivative in the distribution sense?

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SUMMARY

The discussion clarifies that if a function ##u## belongs to the space ##L^2(\mathbb{R})## and satisfies the condition that for all smooth functions ##\psi## in the space ##\mathcal{C}_c^1(\mathbb{R})##, the integral relationship holds, then ##u## is recognized as the derivative of the function ##\varphi## in the distributional sense. This definition is crucial for understanding derivatives within the framework of distribution theory.

PREREQUISITES
  • Understanding of function spaces, specifically ##L^2(\mathbb{R})##.
  • Familiarity with smooth functions and the space ##\mathcal{C}_c^1(\mathbb{R})##.
  • Knowledge of integral calculus and properties of integrals.
  • Basic concepts of distribution theory and its applications.
NEXT STEPS
  • Study the properties of distribution derivatives in more depth.
  • Learn about Sobolev spaces and their relation to distribution theory.
  • Explore the applications of distributions in solving differential equations.
  • Investigate the role of test functions in distribution theory.
USEFUL FOR

Mathematicians, physicists, and students studying advanced calculus or functional analysis, particularly those interested in distribution theory and its applications in various fields.

pellman
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Never mind. I got this one. Couldn't figure out how to delete the post though.
 

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Let ##u\in L^2(\mathbb{R})## be a function such that for all smooth ##\psi\in \mathcal{C}_c^1(\mathbb{R})##, we have that
\int_{-\infty}^{+\infty} \psi(x) u(x)dx = -\int_{-\infty}^{+\infty} \psi'(x) \varphi(x)dx
Then ##u## is said to be the derivative of ##\varphi## in the distributional sense.
 
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micromass said:
Let ##u\in L^2(\mathbb{R})## be a function such that for all smooth ##\psi\in \mathcal{C}_c^1(\mathbb{R})##, we have that
\int_{-\infty}^{+\infty} \psi(x) u(x)dx = -\int_{-\infty}^{+\infty} \psi'(x) \varphi(x)dx
Then ##u## is said to be the derivative of ##\varphi## in the distributional sense.

Thanks, micromass! I had just found the answer and was editing the OP. Still much appreciated though.
 

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