# Looks obvious…but how can i prove this uniformly continuousness?

Let $F(x)$ be a continuous function defined in $\mathbb R$ and this satisfies: $$F(0)>0 \quad \text{and} \quad \text{if} \quad |x|>1 \quad \text{then} \quad F(x)=0$$

I can image that this is uniformly continuous but how can I prove it? Thanks in advance!

• Is $F$ continuous? – CL. Jun 3 '17 at 7:57
• yes, sorry I forgot it so I added it – lacm Jun 3 '17 at 7:59

$F$ is continuous in the compact $[-1,1]$, so it's uniformly continuous on $[-1,1]$. Using that it is easy to prove that given $\varepsilon>0$, there exists $\delta>0$ so that $|x-y|<\delta \Rightarrow |F(x)-F(y)|<\varepsilon$.

• If $x,y\in[-1,1]$, then we have a $\delta_1$ so that $$|x-y|<\delta_1 \Rightarrow |F(x)-F(y)|<\varepsilon$$
• If $x,y\notin[-1,1]$, then it is obvious that for every $\delta$ the condition $$|x-y|<\delta \Rightarrow |F(x)-F(y)|=0<\varepsilon$$ is satisfied.
• If $x\in[-1,1], y\notin[-1,1]$, then we have to find a $\delta_2$ so that $$|x-y|<\delta_2 \Rightarrow |F(x)|<\varepsilon$$ Suppose we have $x<1<y$. Then we have: $$|x-y|\leq |x-1|+|y-1|$$ and $|F(x)|=|F(x)-F(1)|$, because $F(1)=0$ by continuity. So it is enough to pick $\delta_1$, as: $$|x-y|\leq \delta_1 \Rightarrow |x-1|\leq \delta_1 \Rightarrow |F(x)|=|F(x)-F(1)|<\varepsilon$$ A similar reasoning can be done if $x>-1>y$.

So in the end, the choice of $\delta$ is $\delta_1$ given by the uniform continuity of $F$ in $[-1,1]$.

Since $F$ is continuous on the compact set $[-1,1]$, then by the Heine-Cantor theorem it is also uniformly continuous on the same set.

Hence, for every $\epsilon > 0$, there exists $\delta > 0$ such that $$|F(x) - F(y)| < \epsilon \qquad \forall x,y\in [-1,1],\ |x-y| < \delta.$$ Now, it is easy to verify that $$|F(x) - F(y)| < \epsilon \qquad \forall x,y\in \mathbb{R},\ |x-y| < \delta.$$

• Thanks. But I wonder this is correct or not: if a function defined in $\mathbb R$ is uniformly continuous in $(-\infty , a]$ and $[a, \infty)$ then is it uniformly continuous in $R$? – lacm Jun 3 '17 at 8:05
• Yes, it is. (It is important that the two intervals overlap in at least one point.) If you have already proved this result, then the uniform continuity of the function of the exercise follows from its uniform continuity on the intervals $(-\infty, -1]$, $[-1,1]$ and $[1,+\infty)$. – Rigel Jun 3 '17 at 8:23
• Thank you, I'll try it – lacm Jun 3 '17 at 8:34

Hint. Note that a more general statement holds.

If $f$ is continuous in $\mathbb{R}$, $g$ and $h$ are uniformly continuous in $\mathbb{R}$ with $\lim_{x\to +\infty}(f(x)-g(x))=0$ and $\lim_{x\to -\infty}(f(x)-h(x))=0$ then $f$ is uniformly continuous in $\mathbb{R}$.

Take a look here:Prove that the function is uniformly continuous