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Let $I \subseteq \Bbb R$ be non-empty open interval. Let $f : I \rightarrow \Bbb R$ be a real analytic function. Let $y_0 \in I$ be a point such that $f^{\prime}(y_0) \neq 0$

(a) Show that there is an open interval $J \subseteq I$ containging $y_0$ such that $f^{\prime}(x) \neq 0$, $\forall x \in J$

(b)Show that there is an open interval $J^{\prime} \subseteq I$ containing $y_0$ and a constant $C>0$ such that

$$|f(x)-f(y_0)|^{\frac{1}{2}} \le C|f^{\prime}(x)|, \forall x \in J^{\prime}$$

My trial : I solved the problem (a). It suffices to show that $f^{\prime}$ is continuous on the interval $I$. Then, It was induced by the fact that $f^{\prime}$ is uniformly continuous on the interval.

When it comes to a problem (b), I was stuck in here. I don't know how to start to show this inequality. I guessed the result of (a) and Mean Value Theorem would be necessary to solve, but I have no idea of how to apply them. Could you give me a few hint or concept to solve this problem..? Thank you!

fivestar
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1 Answers1

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You can choose $J '$ and $r>0$ such that $|f'(y)| >r$ for all $y \in J'$. Note that $|f'(s)| \leq M|f'(x)|$ on $J'$ if $M =\frac 1 r \sup \{|f'(s)|: s \in J'\}$. Now apply MVT. There exists $t$ between $x$ and $y_0$ such that $|f(x)-f(y_0)|= |f'(t)| |x-y_0| \leq \frac 1 r|f'(t)|^{2} (|x|+|y_0|) \leq \frac 1 r M^{2} |f'(x)|^{2} (|x|+|y_0|) $. Now use that fact that $|x|$ is bounded on $J'$. Taking square roots on both sides we get the desired inequality.

Kavi Rama Murthy
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