Graduate Vector analysis and distributions

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The discussion centers on the vector analysis of the function 1/r, noting that while it is often stated that the Laplacian of 1/r is zero for r ≠ 0, the correct expression is Δ(1/r) = -4πδ(𝑟) when considering the origin. The divergence of the vector field r/r^3 is examined, revealing that it diverges at the origin and yields -2r^(-3) through direct calculation. The divergence theorem is employed to analyze the delta distribution around the origin, indicating that the surface integral remains constant regardless of the chosen surface. This consistency suggests a delta distribution at the origin, which can be rigorously defined based on the chosen interpretation of the delta function. The conversation emphasizes the importance of careful treatment of singularities in vector calculus.
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In many books it is just written that ##\Delta(\frac{1}{r})=0##. However it is only the case when ##r \neq 0##. In general case ##\Delta(\frac{1}{r})=-4\pi \delta(\vec{r})##. What abot ##\mbox{div}(\frac{\vec{r}}{r^3})##? What is that in case where we include also point ##0##?
 
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Direct calculation gives ##-2r^{-3}## if I do it right. it diverges at the Origin.
 
You can use the divergence theorem to compute what delta distribution it is around the origin. When you compute the surface integral you get a constant value no matter which surface you use (you can verify this with spheres pretty easily) and that implies a delta distribution at the origin. You can make this more rigorous depending on how you define a delta distribution
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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