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$g$ is a continous function on $\mathbb{R}$ and $g(0)=0$, $f$ satisfies $|f(x)-f(y)|\le g(x-y)$, I need to show $f$ is continous.

well I tried like this: $|f(x)-f(0)|\le g(x), |f(0)-f(y)|\le g(-y)$,

$|f(x)-f(y)|=|f(x)-f(0)+f(0)-f(y)|\le g(x)+g(-y)$

I can not proceed further, please help.

Myshkin
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  • You want: when $|x - y|$ is small, then $|f(x) - f(y)|$ is small. You have: when $|z| = |z-0|$ is small, then $|g(z)| = |g(z) - g(0)|$ is small. Does that help? – Mees de Vries Apr 07 '17 at 14:18

3 Answers3

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Let $c>0$, since $g$ is continuous at $0$, there exists $d$ such that $|y|<d$ implies that $|g(y)-g(0)|=|g(y)|<c$. Let $x$ be an element of $\mathbb{R}$ and $y$ an element such that $|x-y|<d$, we deduce that $|f(x)-f(y)|<|g(x-y)|<c$. This implies that that $f$ is continuous.

Tsemo Aristide
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Actually we have stronger results, that $f$ is not only continuous, but uniformly continuous.

Definition of continuous: $\forall y$ in the domain of $f$, and $\forall \epsilon>0, \,\exists \delta_{y,\epsilon}$, s.t. $|f(x) - f(y)| <\epsilon$ whenever $|x - y| < \delta_{y,\epsilon}$

Definition of function uniformly continuous: $\forall \epsilon>0, \,\exists \delta_{\epsilon}$, s.t. $|f(x) - f(y)| <\epsilon$ whenever $|x - y| < \delta_{\epsilon}$

Proof. given any $\epsilon_0 > 0$

Since $g$ is continuous, and $g(0) = 0$, we could find a $\delta_{\epsilon_0}$, s.t. $|g(x - y) - g(0)| = g(x-y)< \epsilon_0$, whenever $|x-y - 0| < \delta_{\epsilon_0}$

Thus we have $|f(x) - f(y)| < g(x-y) < \epsilon_0$, whenever $|x-y| < \delta_{\epsilon_0}$

So we find a $\delta$ for each $\epsilon$ uniformly over the domain of $f$, and thus $f$ is uniformly continuous.

Community
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Jay Zha
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Regards Urgent..

To show that the function $f(x)$ is continuous in $\mathbb{R}$, is as same as to show that $$ \lim_{x \rightarrow a} f(x) = f(a), \: \: \: \text{for all } a \text{ in } \mathbb{R}$$

Now to show this, we have info that $g$ is continuous, $g(0)=0 $, and the relation below holds

$$ |f(x)-f(a)| \le g(x-a)$$

From the relation we get

$$ - g(x-a) \le f(x)-f(a) \le g(x-a)$$

Then taking the limit $x \rightarrow a$ , the $g$ term will be $0$ because it is continuous, and then by sandwich we would conclude that $f(x)$ is continuous. $$ 0 \le \lim_{x \rightarrow a} f(x)-f(a) \le 0 $$ $$ \lim_{x \rightarrow a} f(x)-f(a) = 0 $$

Thanks.

Redsbefall
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