Suppose that \(f:\mathbb{R}\to\mathbb{R}\) satisfies \(f(0)=0\). Prove that \(f\) is differentiable at \(x=0\) if and only if there is a function \(g:\mathbb{R}\to\mathbb{R}\) which is continuous at \(x=0\) and satisfies \(f(x)=xg(x)\) for all \(x\in\mathbb{R}\).
Solution
1. (\(\Leftarrow\)) If \(f(x) = xg(x)\) with \(g\) continuous at \(0\), then \(f\) is differentiable at \(0\) with \(f'(0) = g(0)\).
\(\frac{f(x) - f(0)}{x - 0} = \frac{xg(x)}{x} = g(x) \to g(0)\) as \(x \to 0\) (continuity of \(g\) at \(0\); also \(f(0) = 0\cdot g(0) = 0\)).
2. (\(\Rightarrow\)) If \(f\) is differentiable at \(0\) (and \(f(0) = 0\)), define \(g(x) = \frac{f(x)}{x}\) for \(x \neq 0\) and \(g(0) = f'(0)\).
then \(f(x) = xg(x)\) for all \(x\) (for \(x = 0\) both sides are \(0\)).
3. \(g\) is continuous at \(0\): \(\lim_{x \to 0} g(x) = \lim_{x \to 0}\frac{f(x)}{x} = \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} = f'(0) = g(0)\).
2 and the definition of the derivative (using \(f(0) = 0\)).
4. Q.E.D.
1 and 3 give both directions.