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Let $ f: X \subset \mathbb{R}^p \to \mathbb{R}^q $ and $ a \in X$. Supose that for all $ \epsilon > 0 $ exists $ g: X \to \mathbb{R}^q $ continuous at $a$ such as $ \| f(x) - g(x) \| < \epsilon $ for all $x \in X$, prove that $f$ is continuous in $a$.

Demonstration:

Since $g$ is continuous we have:

$ \forall \epsilon > 0 $, $ \exists \delta > 0 $; $ \| x -a \ < \delta \implies \| g(x) - g(a) \| < \epsilon $.

We know by hypothesis that $ \| f(x) - g(x) \| < \epsilon \; \forall \; x \in X $.

We wanto to prove:

$ \forall \epsilon > 0 $, $ \exists \delta > 0 $; $ \| x -a \| < \delta \implies \| f(x) - f(a) \| < \epsilon $.

My intuition was that the norm/distance (in the metric space ) from $ \| f(a) - f(x) \| \le \| g(x) - g(a) \| - \frac{\epsilon}{2} = r $. But I could not follow up since the distance do no implies that $f(x)$ is inside the ball on $g(x)$

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Let $\varepsilon > 0$. By hypothesis, there exists a continuous function $g$ such that $\|f(x) - g(x)\| < \varepsilon$. Since $g$ is continuous at $a$, there exists $\delta > 0$ such that for all $x$, $\|x - a\| < \delta$ implies $\|g(x) - g(a)\| < \varepsilon$. So for all $x$, $\|x - a\| < \delta$ implies $$\|f(x) - f(a)\| \le \|f(x) - g(x)\| + \|g(x) - g(a)\| + \|g(a) - f(a)\| < 3\varepsilon.$$ Since $\varepsilon$ was arbitrary, $f$ is continuous at $a$.

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  • $\begingroup$ THank you for the reply, but how can you affirm that $ \| g(a) - f(a) || < \epsilon $ for some epsilon? $\endgroup$ – Lin Feb 1 '15 at 21:45
  • $\begingroup$ I got it. $ \| f(x) - g(x) \| < \epsilon ; \; \forall x \in X $ in particular $ x=a $ $\endgroup$ – Lin Feb 1 '15 at 21:47
  • $\begingroup$ Yes, that's correct. $\endgroup$ – kobe Feb 1 '15 at 21:48

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