is there any real-valued function with following property

Question

suppose we have a convex continuous function from$\mathbb{R}$ to $\mathbb{R^+}$ $f:\mathbb{R}\to\mathbb{R^+}$ such that
$f(-x)=f(x),lim_{x\to 0}\frac{f(x)}{x}=0 \text{ and } lim_{x\to \infty}\frac{f(x)}{x}=\infty$ and
define $f_{1}(x)$=$e^{(\frac{-1}{x^2})}f(x)$
My question is can we define $f(x)$ such that
$f_1(x) =$ $\begin{cases} 0 & :x\in[-k,k],\text{where}\hspace{0.2cm} k\in \mathbb{R}\\ \text{it will increase } & : x\in(-\infty,-k)\cup(k,\infty) \end{cases} $
I hope I have written my question so that people can understand what I want to say, if there is any mistake please let me know.All I want is a function which is zero in a closed interval and then it increases and the final plot is like convex function,if I keep on changing the value of $k$ it should be zero in all those $[-k,k]$ for every $k\in \mathbb{R}$

Answer 1

what's wrong with:

$f(x)=0, -k\leq x\leq k\\ f(x)=x^2-k^2,(x>k\lor x<-k)$

it is continuous at $k,-k$, it convex and the limits exist at $0,\infty$.

Answer 2

Such a function cannot exist, unless it is identically $=0$. To see this, assume $f_1$ is increasing on $(-\infty, -k)$ and not $=0$. Since $f$ is continuous, so is $f_1$, and this means that $f_1(x) \le 0 $ for $x<-k$, and, if not identically $0$, $f_1(x) < 0$ for some $x< -k$.

Since both $f$ and $e^{-\frac{1}{x^2}}$ are non negative this is a contradiction.