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The problem asks to prove that if $g: \mathbb{R} \rightarrow \mathbb{R} \in C^1$ then $g$ is stable.

By definition of stability of a function on a point

$g:\mathbb{Y} \rightarrow \mathbb{X} $ is stable in $y_0 \implies \exists$ $ \delta, M > 0 :$ $ sup \frac{||g(y) - g(y_{0}||}{||y-y_{0}||} < M $ with $y \in \mathbb{Y} \text{ , } 0 < ||y-y_{0}|| < \delta$

I have thought in the mean value theorem that relates the derivate of a function with a line similar to what I have hear but don't know exactly how to do it. Any hint or help?

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    How does $M$ play into all of this? – 2017-02-24
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    @AdamHughes fixed, sorry forgot M at the time of write the formula – 2017-02-24
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    Take any $\delta > 0$. The value under the supremum for any $y$ with $0 < \Vert y-y_0\Vert < \delta$ is then equal, as you stated, to some $g'(w)$ with $w$ between $y$ and $y_0$. Can you give an interval (not using $y$) in which $w$ must lie? Then, taking into account that $g$ has a continuous derivative, can you come to the conclusion you need? – 2017-02-24

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Take any $y_0$, choose any $\delta > 0$ you want. Since $g \in C^1$, $g'(y)$ exists and is continuous in the closed interval $[y_0-\delta,y_0+\delta]$. Hence, $g'(y)$ is bounded on that interval: $\exists M:|g'(t)| < M$ for $t \in [y_0-\delta,y_0+\delta]$.

For any $ y \in [y_0-\delta,y_0+\delta], y \ne y_0$ we also have $$\frac{|g(y)-g(y_0)|}{|y-y_0|} = g'(\xi)$$

for some $\xi \in (\min (y,y_0), \max (y,y_0))$ due to the intermediate value theorem. It follows that $\xi \in [y_0-\delta,y_0+\delta]$, hence we know that

$$\frac{|g(y)-g(y_0)|}{|y-y_0|} < M$$.

From here the conclusion

$$\sup \frac{|g(y)-g(y_0)|}{|y-y_0|} \le M$$ for all $0 < |y-y_0| < \delta$ follows.

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    oh sorry, forgot to the you that you comment helped me :) . Anyway thank you for posting an answer. – 2017-02-27