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This question is inspired by the accepted answer found here.

Suppose we have a sequence of functions say $$f(y)\cdot\frac{g(x-y+h)-g(x-y)}{h}$$ where we are assuming that $f$ is Lebesgue integrable and $g$ is differentiable and bounded. By specific question is how are they viewing the function above? As a function of $x$ or a function of $y$? Suppose it is a function of $y$, and take the limit when $h$ tends to $0$. In which we can we would get that $$f(y) \cdot \lim _{h \rightarrow 0} \frac{g(x-y+h)-g(x-y)}{h}.$$ Here is my confusion, this is of course the definition of the derivative, but what is it with respect to? The variable $x$ or $y$? If we view it as a function of $y$, then $x$ is presumed to be constant, in which case I am tempted to say that the above equals $f(y) \frac{d}{dy}g(x-y)$ However, I do not think this is what they use in the proof.

Hrafn Magnus
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1 Answers1

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It is with respect to the only variable of $g$ say $z$ for $g(z)$, but evaluated at $x-y$, so $g'(x-y)$. We would also denote this as $$\left.\frac{d}{dz}\right\rvert_{x-y} g(z).$$

Note that we could denote it as $$\left.\frac{d}{dx}\right\rvert_{x-y} g(x)$$ but it is a case of bad notation since $x$ appears in two places with completely different meanings. It is the same problem as the notation $$f(x)=\int_{0}^{x}g(x)dx.$$

spinosarus123
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