3

Hi everyone I'm asking two thinks is this proof correct (my other idea was using limit of sequences)? and are there a simpler alternative than this using Newton's approximation? If someone could help me I'd be so thankful.

Proposition: Let $X,Y$ be subsets of $\mathbb{R}$, $f:X\rightarrow Y$ and $g:Y\rightarrow \mathbb{R}$ be functions, $x_0\in X$, $y_0\in Y$ are limit points of $X$ and $Y$ respectively, and $y_0=f(x_0)$. Then if $f$ is differentiable at $x_0$ and $g$ is differentiable at $y_0$, then the composition is differentiable at $x_0$ and $(g\circ f)'(x_0)=g'(y_0)f'(x_0)$, i.e.,

$$\lim_{x\rightarrow x_0;x\in X\backslash\{x_0\}} \frac{(g\circ f)(x)-(g\circ f)(x_0)}{x-x_0}=g'(y_0)f'(x_0)$$

Proof: It will suffice to show that given a $\varepsilon>0$ there exists a $\delta>0$ such that $(g\circ f)(x)$ is $\varepsilon |x-x_0|$-close to $(g\circ f)(x_0)+g'(y_0)f'(x_0)(x-x_0)$ whenever $x\in X$ and $x$ is $\delta$-close to $x_0$. After a little algebra this is equivalent to

$g(f(x))-g(f(x_0))-g'(y_0)f'(x_0)(x-x_0)=$ \begin{align} \big[g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))\big] +\big[ g'(f(x_0))(\,(f(x)-f(x_0)-f'(x_0)(x-x_0))\,) \big]\end{align}

Since $\big|f(x)-f(x_0)-f'(x_0)(x-x_0) \big|\le |x-x_0| \varepsilon_1$ because $f$ is differentiable at $x_0$ by hypothesis. We set $\varepsilon_1=\varepsilon/2(| g'(f(x_0))|+1)$ and $\delta_1>0$ such that for all $\{x\in X: |x-x_0|\le \delta_1\}$ we have

$$\big|f(x)-f(x_0)-f'(x_0)(x-x_0) \big|\le \frac{ |x-x_0| \varepsilon}{2(| g'(f(x_0))|+1)}$$

Now let $\varepsilon_2=\varepsilon/2(|f'(x_0)|+1))$. Since $|g(y)-g(y_0)-g'(y_0)(y-y_0)|\le |y -y_0|\varepsilon_2$ we can find some $r>0$ such that whenever $|y-y_0|\le r$ the above inequality holds. Since $f$ is differentiable it follows that is continuous at $x_0$ then there is $\delta_2$, $0<\delta_2\le \delta_1$, such that $|f(x)-f(x_0)|\le r$ and $|f(x)-f(x_0)|\le |x-x_0|(|f'(x_0)|+1)$ at the same time. Thus we have

\begin{align}|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\le \frac{|f(x) -f(x_0)|\varepsilon}{2(|f'(x_0)|+1)}\\ \le|x-x_0|\varepsilon/2 \end{align}

For instance we have

$$\begin{align}|g(f(x))-g(f(x_0))-g'(y_0)f'(x_0)(x-x_0)|\le |x-x_0|\varepsilon /2+ |g'(f(x_0))|\frac{ |x-x_0| \varepsilon}{2(| g'(f(x_0))|+1)} \\ \le |x-x_0| \varepsilon\end{align}$$

whenever $x$ is $\delta_2$-close to $x_0$ as claimed.

Jose Antonio
  • 7,524

1 Answers1

2

An idea for a much shorter and simpler proof using continuity, and using your same notation:

$$g\circ f(x_0):=\lim_{x\to x_0}\frac{g\circ f(x)-g\circ f(x_0)}{x-x_0}=$$

$$=\lim_{x\to x_0}\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}\cdot\frac{f(x)-f(x_0)}{x-x_0}=g'(f(x_0))\cdot f'(x_0)$$

since, by continuity, we have that

$$x\to x_0\implies f(x)\to f(x_0)$$

DonAntonio
  • 215,504
  • After all thanks. But what garantee us that $f(x)-f(x_0)\not= 0$ for example for a constant function? – Jose Antonio Feb 26 '14 at 19:00
  • Well @JoseAntonio, the case of a constant function can easily be dealt with separatedly... – DonAntonio Feb 26 '14 at 19:05
  • What if $f$ is non-constant, but $f(x)=f(x_0)$ for at least one $x$ in every $\varepsilon$ neighborhood of $x_0$? Then the limit does not hold. – Tom Jan 11 '19 at 01:15
  • @Tom This is oooooold...but anyway: if $;f;$ isn't constant in some neighborhood of $;x_0;$ (and this is the basic, hidden assumption), then there exists a subneighborhood for which $;f(x)\neq f(x_0);$, by continuity of $;f;$ at $;x_0;$ . This is exactly the same as saying: if $;f;$ continuous at $;x_1;$ and $;f(x_1)>0;$ then $;f(x)>0;$ for all $;x;$ in some nieghborhood of $;x_1;$ . – DonAntonio Jan 11 '19 at 10:00
  • Being nonconstant in a neighborhood is not enough, the function could also take on $f(x_0)$ infinitely many times in every neighborhood of $x_0$, for example, $x\sin(\frac{1}{x})$ at 0. In order to prove, you must expose all hidden assumptions! – Tom Jan 15 '19 at 19:51
  • Sure, lets say $f(x)=x\sin(\frac{1}{x})$ when $x\not= 0$ and $0$ otherwise. $f$ is continuous at zero. Now let $g(x)=x$. Say we compose $f\circ g$. Do you see the problem now? – Tom Jan 15 '19 at 20:03
  • Then let $h(x)= f(x)+1$, now consider $h\circ g$. Do you understand? Being non-constant, non-zero and continuous are not enough. – Tom Jan 15 '19 at 20:31
  • @Tom I really don't get what is your point: that $;h;$ isn't continuous at zero...let's make some order in your original question: it is implicitly assumed that $;f;$ isn't constant in some neighborhood of $;x_0;$ (otherwise there's hardly anything to prove), and from here that there exists (perhaps another) neighborhood of $;x_0;$ where $;f(x)\neq f(x_0);$ in that (perhaps other) neighborhood...there you go! Why? Because $;f;$ is continuous at $;x_0;$ . Period. – DonAntonio Jan 15 '19 at 20:43
  • $g$ must be differentiable at $x_0$ and $f$ must be differentiable at $g(x_0)$. – Tom Jan 15 '19 at 20:48
  • @Tom Ah, now I see my flaw: I interchanged (now, 5 years ago I've no idea what I was thinking of) the roles of $;f,g;$ in my mind (as I'm used more to $;f\circ g;$ than the other way around). Ugly mistake...thanks, this discussion stays in order to help others to see the flaw in my answer and I shall delete some of my comments above in order not to confuse those potential future reads. – DonAntonio Jan 15 '19 at 20:58