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evinda
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Hey! ![Cool :cool: :cool:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I am looking at the following exercise:
Let $f_n: I \to \mathbb{R}$ a sequence of functions,that do not get zero anywhere.We suppose that $f_n \to f$ uniformly and that there is a $M>0$ such that $f(x) \geq M, \forall x \in I$.Show that $\frac{1}{x_n} \to \frac{1}{f}$ uniformly.
That's what I have tried:
Let $\epsilon>0.$
Since $f_n \to f$ uniformly $\exists n_0$ such that $\forall n \geq n_0: sup_{x \in I} \{f_n-f\} \leq \epsilon $.
For these $n \text{ and } \forall x \in I$, we have:
$$|\frac{1}{f_n(x)}-\frac{1}{f(x)}|=|\frac{f(x)-f_n(x)}{f_n(x) \cdot f(x)}| \leq \frac{sup_{x \in I} \{f_n-f\}}{|f_n(x)| |f(x)|}$$
$$M \leq |f(x)|=|f(x)-f_n(x)+f_n(x)| \leq |f_n(x)-f(x)|+|f_n(x)| \leq \epsilon + |f_n(x)| \Rightarrow |f_n(x)| \geq M-\epsilon \Rightarrow \frac{1}{|f_n(x)|} \leq \frac{1}{M-\epsilon}$$
So, $$sup_{x \in I} \{|\frac{1}{f_n(x)}-\frac{1}{f(x)}|\} \leq \frac{\epsilon}{M \cdot (M- \epsilon)} \to 0 $$
So, $$\frac{1}{f_n} \to \frac{1}{f} \text{ uniformly }$$
Could you tell me if it is right? Do I have to take maybe a specific $\epsilon$? (Thinking)
I am looking at the following exercise:
Let $f_n: I \to \mathbb{R}$ a sequence of functions,that do not get zero anywhere.We suppose that $f_n \to f$ uniformly and that there is a $M>0$ such that $f(x) \geq M, \forall x \in I$.Show that $\frac{1}{x_n} \to \frac{1}{f}$ uniformly.
That's what I have tried:
Let $\epsilon>0.$
Since $f_n \to f$ uniformly $\exists n_0$ such that $\forall n \geq n_0: sup_{x \in I} \{f_n-f\} \leq \epsilon $.
For these $n \text{ and } \forall x \in I$, we have:
$$|\frac{1}{f_n(x)}-\frac{1}{f(x)}|=|\frac{f(x)-f_n(x)}{f_n(x) \cdot f(x)}| \leq \frac{sup_{x \in I} \{f_n-f\}}{|f_n(x)| |f(x)|}$$
$$M \leq |f(x)|=|f(x)-f_n(x)+f_n(x)| \leq |f_n(x)-f(x)|+|f_n(x)| \leq \epsilon + |f_n(x)| \Rightarrow |f_n(x)| \geq M-\epsilon \Rightarrow \frac{1}{|f_n(x)|} \leq \frac{1}{M-\epsilon}$$
So, $$sup_{x \in I} \{|\frac{1}{f_n(x)}-\frac{1}{f(x)}|\} \leq \frac{\epsilon}{M \cdot (M- \epsilon)} \to 0 $$
So, $$\frac{1}{f_n} \to \frac{1}{f} \text{ uniformly }$$
Could you tell me if it is right? Do I have to take maybe a specific $\epsilon$? (Thinking)