Let $f : [0, \infty) \to \mathbb{R}$ be differentiable on $(0,\infty)$ and $f(0+) \in \mathbb{R}$ exists (not necessarily equal to $f(0)$). Typically, the right derivative of $f$ at $0$ is defined as $\lim_{h \to 0+} \frac{f(h) - f(0)}{h}$ if it exists. However, I am concerned with $\lim_{h \to 0+} \frac{f(h) - f(0+)}{h}$. Is this limit common and has some special name? For me, it seems to be some kind of "derivative to the boundary" of the function $f$ with restricted open domain $(0,\infty)$.
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http://math.stackexchange.com/questions/265552/a-notation-question-what-does-it-mean-gx-fx Folland uses the notation $G(x) = F(x+) = \lim\limits_{y\to x^+} F(y)$ which finds some use when proving monotone functions are Lebesgue differentiable $m$-almost everywhere. $G'(x)$ would be precisely the limit you are interested in if I'm not mistaken. I don't know that it has a special name however, someone else might be able to elaborate. It can be shown as well that $G'(x) = F'(x)~~m$-almost everywhere. – JMoravitz Nov 21 '14 at 08:56
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I know the notation ... I edit the question above. – yada Nov 21 '14 at 09:07
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If $f$ is continuous at $0$ then $f(0)=f(0+)$ and both definitions are the same. If $f(0)\ne f(0+)$, then $f$ is discntinuous at $0$ and $\lim_{h\to0^+}(f(h)-f(0))/h$ does not exist. In this case the right concept of derivative would be $\lim_{h\to0^+}(f(h)-f(0+))/h$. Another way to think of this is the following: redefine $f$ at $x=0$ so that it becomes continuous, that is, redefine $f(0)=f(0+)$.
Julián Aguirre
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Yes, so I think the right name would be something like "right derivative" ... it generalizes the definition of a right-derivative for continuous functions like $f(x) = |x|$ where $\lim_{h \to 0+} (f(h) - f(0))/h$ exists and is equal to $\lim_{h \to 0+} (f(h) - f(0+))/h$ since $f(0+) = f(0)$. – yada Nov 22 '14 at 10:20