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## Main Question or Discussion Point

Hello,

I was trying to understand Green's function and I stumbled across the following statements which is confusing to me.

I was referring to the following site

http://www.math.ohio-state.edu/~gerlach/math/BVtypset/node79.html [Broken]

Here the author says the following

"What if $ u$ is not a continuously differentiable function? Then its image $ Lu$ is not square-integrable, but the inner product <v, Lu> is still well-defined because it is finite. For example, if u is a function which has a kink, then $ Lu$ would not be defined at that point and $ Lu$ would not be square-integrable. Nevertheless, the integral of $ \overline v Lu$ would be perfectly finite."

I dont understand the fact is if Lu is not defined how can u define an inner product with v at any point, ie <v, Lu>. What does it mean physically at all, is it a mathematical jugglery to move the L operator to v and then say that look it is still defined? I am totally confused.

Thanks a lot for any help in advance.

I was trying to understand Green's function and I stumbled across the following statements which is confusing to me.

I was referring to the following site

http://www.math.ohio-state.edu/~gerlach/math/BVtypset/node79.html [Broken]

Here the author says the following

"What if $ u$ is not a continuously differentiable function? Then its image $ Lu$ is not square-integrable, but the inner product <v, Lu> is still well-defined because it is finite. For example, if u is a function which has a kink, then $ Lu$ would not be defined at that point and $ Lu$ would not be square-integrable. Nevertheless, the integral of $ \overline v Lu$ would be perfectly finite."

I dont understand the fact is if Lu is not defined how can u define an inner product with v at any point, ie <v, Lu>. What does it mean physically at all, is it a mathematical jugglery to move the L operator to v and then say that look it is still defined? I am totally confused.

Thanks a lot for any help in advance.

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