Prueba existe de que $f'(0)$ y $f'(0) = b/(a - 1)$

Question

<blockquote> <p><strong>Problema</strong>:<br>Si $f(x)$ es continua en $x=0$ y $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ son constantes y $|a|>1$, prueba existe de que $f'(0)$ y $f'(0)=\dfrac{b}{a-1}$.</p> </blockquote> <p>Este enfoque está definitivamente mal:</p> <p>\begin{align} b&=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &=af'(0)-f'(0)\\ &=(a-1)f'(0) \end {Alinee el}</p> <p>Le mostraré un caso por qué este enfoque es incorrecto:</p> <blockquote> <p> $$f(x) = \begin{cases} 1,&x\neq0\\ 0,&x=0 \end{casos}$$ $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$<br>$\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$, $\lim_{x\to0}\dfrac{f(x)}{x}=\infty$</p> </blockquote> <p>¿Alguien sabe cómo demostrarlo? ¡Gracias de antemano!</p>

Answer 1

Esta es una pregunta difícil y la solución es algo no evidente. Sabemos que $$\lim_{x \to 0}\frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) \a 0$ as $x \to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, \ldots, n$ we get $$f(x) - f(x/a^{n}) = bx\sum_{k = 1}^{n}\frac{1}{a^{k}} + x\sum_{k = 1}^{n}\frac{g(x/a^{k})}{a^{k}}$$ Letting $n \to \infty$ and using sum of infinite GP (remember it converges because $|a| > 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = \frac{bx}{a - 1} + x\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x \a 0$ we get $$f'(0) = \lim_{x \to 0}\frac{f(x) - f(0)}{x} = \frac{b}{a - 1} + \lim_{x \to 0}\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$

La suma $$\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x \to 0$ because $g(x) \a 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)\a 0$ as $x \to 0$, it follows that for any $\epsilon > 0$ there is a $\delta > 0$ such that $|g(x)| < \epsilon$ for all $x$ with $0 <|x| < \delta$. Since $|a| > 1$ it follows that $|x/^{k}| < \delta$ if $|x| < \delta$ and therefore $|g(x/a^{k})| < \epsilon$. Thus if $0 < |x| < \delta$ we have $$\left|\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}\right| < \sum_{k = 1}^{\infty}\frac{\epsilon}{|a|^{k}} = \frac{\epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x \to 0$.

Por lo tanto $f'(0) = b/(a - 1)$.

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Por CIERTO, el resultado de la pregunta se mantiene incluso si $0 < |a| < 1$. Deje $c = 1/a$, de modo que $|c| > 1$. Ahora tenemos $$\lim_{x \to 0}\frac{f(ax) - f(x)}{x} = b$$ implies that $$\lim_{t \to 0}\frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = \frac{-bc}{c - 1} = \frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $= 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|| a \neq 1$.

Answer 2

Sugerencia: Estás muy cerca.

Escribir la expresión como $$a\frac{f(ax)-f(0)}{ax}-\frac{f(x)-f(0)}{x}$$ Note that $x\to 0$ if and only if $ax\to 0$ (since $a\neq 0$).

¿Se deduce de esto?