If $f:R \to R$ is a continuous function and there are $a \geq 0$ and $b \geq 0$ such that $\|f(x)\| \leq a\|x\| + b$ then $f$ is uniformly continuous.

Question

I must verify if the following statement is true: if $f:R \to R$ is a continuous function and there are $a \geq 0$ and $b \geq 0$ such that $|f(x)| \leq a|x| + b$ then $f$ is uniformly continuous.

My idea:

It's false. Let $b \in R$. We define $f$ recursively: In the interval $[0,1]$, $f$ increases linearly from $0$ to $b$ in $[0,1/2]$ then decreases to $0$ in $(1/2,1]$, put $x_1=1$. Already defined $f$ in $[0,x_n]$, we put $f$ on $[x_n,x_n+1/n]$ also from $0$ increasing to $b$ then decreasing to $0$, linearly, and put $x_{n+1}=x_n+1/n$. Since $\sum 1/n$ diverges, this process define a continuous function $f:[0,\infty] \to R$. Setting $f(x)=-f(x)$ for $x < 0$. $f$ is a continuous function in the hypothesis of the question ($a=0$) that cannot be uniformly continuous since for any $\epsilon < b$ and for any $\delta > 0$, if we take $1/n < \delta$, we have $|\left(x_n+1/n \right) - x_{n+1}| < \delta$ and $|f(x_n+1/n) -f(x_{n+1})| = b > \epsilon $.

Is my idea wrong? Thanks in advance!

Answer 1

If you consider $f(x)=xsin(x)$ then you have that

$|f(x)|\leq |x|$ but in any case $f$ is not uniformly continuos.

In any case your counter-example is correct.