A function $f: {\mathbb{R}^m} \to {\mathbb{R}^n}$ is said to be differentiable at $x$ if there exists a linear function $L(x)$ such that $f(x+h)-f(x)=L(x)h + o(h)$, as $h\to0$.
I'm abit confused about the use of little $o$ here. Which one of the interpretation is correct?
- There exists a function $\alpha(t) = o(t)$ as $t\to 0$, such that for any $h \in \mathbb{R}^m$ $f(x+h)-f(x)=L(x)h + \alpha(h)$. Thus in this interpretation $o(h)$ is an $o(t)$ function evaluated at $h$, and the "addition" here is actual number addition.
- The function $g(h) = f(x+h)-f(x)$ is $L(x) + o(h)$. Thus in this definition, we are actually performing function additions, and $o(h)$ represents an $o(h)$ function rather than some $o(t)$ function evaluated at $h$.
Based on possibly the answer the above question, I'm confused about the use of little o notation in the proof of differentiation of function composition. Here $g$ is differentiable at $f(x)$ and $g'(f(x))$ denote the differential. $$g(f(x + h)) - g(f(x)) = g'(f(x))(f(x + h) - f(x)) + o(f(x + h) - f(x))$$ What does $o(f(x+h)-f(x))$ mean here? Does it mean an $o(t)$ function evaluated at $f(x+h)-f(x)$, or does it mean it is $o(z)$, where $z(h) = f(x+h)-f(x)$. Although by definition of differential, $g(f(x)+k) - g(f(x)) = g'(f(x))k + o(k)$ as $k\to 0$, I feel like it is ambibuguous to just substitue $k$ with any expression.
Everything depends on $x$, so maybe $o_x(h)$ would be more accurate.
Anyway, it is vector addition, not number addition: in $\mathbf R^n$, $f(x+h) - f(x) - L(x)h = o(h)$ as $h \to 0$ in $\mathbf R^m$, i.e., $|\!|f(x+h) - f(x) - L(x)h|\!|/|h| \to 0$ as $h \to 0$. Or $f(x+h) = f(x) + L(x)h + \alpha_x(h)$, where $\alpha_x$ is a function on a small neighborhood of $0$ in $\mathbf R^m$ with values in $\mathbf R^n$ such that $|\!|\alpha_x(h)|\!|/|h| \to 0$ as $h \to 0$.