Hello,(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Non continuously differentiable but inner product finite

**Physics Forums | Science Articles, Homework Help, Discussion**