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I have three functions, $f,g,h$ that all map from $[a,b]$ to $\mathbb{R}$. I know the following:

$$f(x) \leq g(x) \leq h(x)\ \forall x\in [a,b] $$ $$x_0 \in (a,b)$$ $$f(x_0)=h(x_0)$$ $$h'(x_0) \ \& \ f'(x_0) \ exist$$

I want to show that $g'(x_0)$ exists and that $g'(x_0)=f'(x_0)=h'(x_0)$

Any suggestions? I'm really lost. I tried applying the definition of something being differentiable at a point but made no progress.

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Consider the definition:

$g$ is differentiable at $x_0$ if $$ \lim_{x\to x_0}\frac{g(x_0)-g(x)}{x-x_0} $$ exists or equivalent if $$ \liminf_{x\to x_0}\frac{g(x_0)-g(x)}{x-x_0}=\limsup_{x\to x_0}\frac{g(x_0)-g(x)}{x-x_0}. $$ or equivalent if $$ \lim_{\substack{x\to x_0\\x>x_0}} \frac{g(x_0)-g(x)}{x_0-x} \text{ and } \lim_{\substack{x\to x_0\\x<x_0}} \frac{g(x_0)-g(x)}{x_0-x} $$ exist and are equal.

Remark:

The limit $$ \lim_{\substack{x\to x_0\\x>x_0}} \frac{g(x_0)-g(x)}{x_0-x} $$ exists if $$ \liminf_{\substack{x\to x_0\\x>x_0}}\frac{g(x_0)-g(x)}{x-x_0}=\limsup_{\substack{x\to x_0\\x>x_0}}\frac{g(x_0)-g(x)}{x-x_0}. $$ and analogue for $<$ insteat of $>$.

Proof:

Now we have $g(x_0)=f(x_0)=h(x_0)$. If $x<x_0$ we get $$ \frac{h(x_0)-h(x)}{x_0-x}= \frac{g(x_0)-h(x)}{x_0-x}\leq \frac{g(x_0)-g(x)}{x_0-x}= \frac{f(x_0)-g(x)}{x_0-x}\leq \frac{f(x_0)-f(x)}{x_0-x} $$ and for $x>x_0$ we get $$ \frac{h(x_0)-h(x)}{x_0-x}= \frac{g(x_0)-h(x)}{x_0-x}\geq \frac{g(x_0)-g(x)}{x_0-x}= \frac{f(x_0)-g(x)}{x_0-x}\geq \frac{f(x_0)-f(x)}{x_0-x} $$ Since $f$ and $h$ are differentiable at $x_0$ we get \begin{align*} f'(x_0)&=\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}= \lim_{\substack{x\to x_0\\x>x_0}}\frac{f(x_0)-f(x)}{x_0-x} \\&\leq \lim_{\substack{x\to x_0\\x>x_0}}\frac{h(x_0)-h(x)}{x_0-x} =\lim_{x\to x_0}\frac{h(x_0)-h(x)}{x_0-x}=h'(x_0) \end{align*} and \begin{align*} f'(x_0)&=\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}= \lim_{\substack{x\to x_0\\x<x_0}}\frac{f(x_0)-f(x)}{x_0-x} \\&\geq \lim_{\substack{x\to x_0\\x<x_0}}\frac{h(x_0)-h(x)}{x_0-x} =\lim_{x\to x_0}\frac{h(x_0)-h(x)}{x_0-x}=h'(x_0) \end{align*} From $f'(x_0)\leq h'(x_0)$ and $f'(x_0)\geq h'(x_0)$ we get $f'(x_0)=h'(x_0)$ and you can conclude $$ \lim_{x\to x_0} \frac{h(x_0)-h(x)}{x_0-x}=\lim_{x\to x_0} \frac{g(x_0)-g(x)}{x_0-x}=\lim_{x\to x_0} \frac{f(x_0)-f(x)}{x_0-x} $$

If you think, the last argument is to sloppy you can extend the last argument following way:

Since $ \frac{h(x_0)-h(x)}{x_0-x}\leq \frac{g(x_0)-g(x)}{x_0-x}$ for all $x\in(a,x_0)$ you get $$ \liminf_{\substack{x\to x_0\\x<x_0}}\frac{h(x_0)-h(x)}{x_0-x}\leq \liminf_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x} $$ and $\frac{g(x_0)-g(x)}{x_0-x}\leq \frac{f(x_0)-f(x)}{x_0-x}$ for all $x\in(a,x_0)$ imply $$ \limsup_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x}\leq \limsup_{\substack{x\to x_0\\x<x_0}}\frac{f(x_0)-f(x)}{x_0-x} $$ These are valid since $\liminf$ and $\limsup$ can be defined even if the limit doesn't exists or the term is unbounded. And now you can use that $f$ and $g$ are differentiable at $x_0$ which imply $$ h'(x_0)=\lim_{x\to x_0}\frac{h(x_0)-h(x)}{x_0-x}= \liminf_{\substack{x\to x_0\\x<x_0}}\frac{h(x_0)-h(x)}{x_0-x} $$ and $$ f'(x_0)=\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x_0-x}= \limsup_{\substack{x\to x_0\\x<x_0}}\frac{f(x_0)-f(x)}{x_0-x} $$ So we get $$ h'(x_0)\leq \liminf_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x} \leq \limsup_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x}\leq f'(x_0) $$ Finally we use $f'(x_0)=h'(x_0)$ and conclude $$ h'(x_0)=\liminf_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x}= \limsup_{\substack{x\to x_0\\x<x_0}}\frac{g(x_0)-g(x)}{x_0-x}=f'(x_0) $$ Therefore $\lim_{\substack{x\to x_0\\x<x_0}} \frac{g(x_0)-g(x)}{x_0-x}$ exists and equals $f'(x_0)$ and $h'(x_0)$. Analogue we can prove that $\lim_{\substack{x\to x_0\\x>x_0}} \frac{g(x_0)-g(x)}{x_0-x}$ exists and equals $f'(x_0)$ and $h'(x_0)$. Together we get that $g$ is differentiable at $x_0$ and $g'(x_0)$ equals $f'(x_0)$ and $h'(x_0)$.

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  • $\begingroup$ Thanks!! I think I get this now $\endgroup$ – Analysis is fun Nov 12 '17 at 10:05
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    $\begingroup$ Hopefully I fixed now all gaps in the proof. I rewrote some parts, because it was wrong. $\endgroup$ – Mundron Schmidt Nov 12 '17 at 10:27
  • $\begingroup$ You don't really need the limsup and liminf. Using the given conditions one can show that $f'(x_{0})=h'(x_{0})$ and then use Squeeze Theorem to get $g'(x_{0})=f'(x_{0})=h'(x_{0})$. $\endgroup$ – Paramanand Singh Nov 12 '17 at 14:45

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