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The whole question: Given $f: X \rightarrow \mathbb{R}$, suppose that for every $\epsilon>0$ we can obtain some continuous function $g: X \rightarrow \mathbb{R}$ such that $|f(x)-g(x)|<\epsilon$ for any/arbitrary/whatever $x\in X$. Prove that $f$ is also continuous.

What i've tried so far: given $a\in \mathbb{R}$, manipulate

$|f(x)-f(a)| = |f(x)-f(a)+g(a)-g(a)+g(x)-g(x)| \leq$

$|g(a)-f(a)|+|g(x)-g(a)|+|f(x)-g(x)|$,

and, taking some care with $\epsilon$, this would imply $|f(x)-f(a)| $ less than $\epsilon$. But i can't ensure that we are talking about the same $g$, since for every different $x$ it is possible to approximate $f$ with some $g$.

Any hints?

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    You did exactly what I would have done. I think it refers that there is ONE function $g$ that approximates $f$ in all its domain. This is what I understand at least. – 2017-02-04
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    It is the same $g$. Note that the condition says that $|f(x) - g(x)| < \epsilon$ holds for all $x\in X$. – 2017-02-04
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    I don't understand your problem. However, I don't see how are you going to estimate $|g(x)-g(a)|$ term. – 2017-02-04
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    @Wolfram It is specified that $g$ is continuous so as long as $|x-a|$ is small enough this can be made as small as possible. – 2017-02-04
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    But you pick $g$ **after** you pick $x$ and $a$, if I got you right. That means for any $g$ for which other two terms are small $|x-a|$ can be not small enough. – 2017-02-04
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    This question is originally written in portuguese. I made my best to translate it. In portuguese it is written: "qualquer que seja $x\in X$". It means something like "for an arbitrary $x$", "whatever $x$". Is it really the same $g$? – 2017-02-04
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    If it was a different $g$ for all $x$ then $g$ would only be needed at one point so it would make no sense to talk about it being continuous. It is for sure just one continuous $g$ and *for any* means *for all* here. – 2017-02-04
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    You question is fine. The same with what? I don't get it. – 2017-02-04
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    Yes, we are talking about the same $g$. The questions states that $g$ is given when $\epsilon$ is given, so for any $x$ it is true that $|f(x)-g(x)|<\epsilon$. So manipulating like i did kills the problem. I misunderstood that $g$ was given based on $x$. – 2017-02-04

3 Answers 3

0

Let $\epsilon>0$ be given. Hence, there exists some $g: X \rightarrow \mathbb{R}$ continuous such that $|f(x)-g(x)|<\epsilon$ for any $x\in X $. For $a \in X$, since $g$ is continuous, there is some $\delta>0$ such that $x\in X$ and $|x-a|<\delta$ implies $|g(x)-g(a)|<\epsilon/3$. In particular, for $x \in (a-\delta,a +\delta)$, the hypothesis ensures that $|f(x)-g(x)|<\epsilon/3$ and $|f(a)-g(a)|<\epsilon/3$. Thus, for $|x-a|<\delta$:

$|f(x)-g(x)|=|f(x)-f(a)+g(a)-g(a)+g(x)-g(x)|\leq |g(a)-f(a)|+|g(x)-g(a)|+|f(x)-g(x)|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon$

This proves that $f$ is continuous.

4

Let $\epsilon>0$ and choose $g$ such that $|f(x)-g(x)| < {1 \over 3} \epsilon$ for all $x$.

Now choose $\delta$ such that if $x \in B(a , \delta)$ then $|g(x)-g(a)| < {1 \over 3} \epsilon$.

If $x \in B(a,\delta)$ then $|f(x)-f(a)| \le |f(x)-g(x)| + |g(x)-g(a)| + |g(a)-f(a)| < \epsilon$.

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    How come the answer got unaccepted? – 2017-02-07
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For every $c>0$, there exists $g_c$ such that $|g_c(x)-f(x)|0$ such that $|y-z|

For every $y\in [x-e/2,x+e/2]$, $|f(y)-f(x)|\leq |f(y)-g_c(y)|+|g_c(y)-g_c(x)|+|g_c(x)-f(x)|