# Continuity of functions inside a open ball

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 $g(x)$

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