Proving that a function $f(x,y)$, that is unbounded in every direction, is uniformly bounded below by $1$ outside some disc of large enough radius

Question

I have a smooth function $f(x,y)$ that is unbounded in every direction. In other words, if we choose a direction $(a,b)\in S^1$ and keep moving along the curve $(ta,tb)$, then $$\lim_{t\to\infty}f(ta,tb)=\infty.$$

Can we conclude that there exists some disc of radius $R$ such that outside of this disc, the function $f(x,y)> 1$ (alternatively, $f(x,y)\geq 1$; see below)?

Note: The "obvious" way to approach this problem is to define a function $g(\theta)$ on $S^1$, where $$g(\theta)=\inf\limits_{t_0} \{t_0:f(t\theta)\geq 1\text{ for all }t\geq t_0\}.$$ If this is a continuous function, then using the compactness of $S^1$, we can conclude that the function attains a maximum at some point on $S^1$, and then use that maximum value to define $R$.

However, based on a couple of examples that I've constructed, this function is not turning out to be continuous. However, changing the requirement from $f(x,y)>1$ to $f(x,y)\geq 1$ seems to do the trick. Is this true in general?

Answer 1

A counterexample: $f(x,y):=x^2y^2$ if $xy\ne0$ and $f(x,y):=|x|+|y|$ otherwise.

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The OP has changed the question to additionally require that $f$ be smooth. Here is a smooth counterexample: $f(x,y):=x^2y^2+(x^2+y^2)e^{-x^2y^2(x^2+y^2)}$.

Indeed, take any $(a,b)\in S^1$. If $ab\ne0$, then $f(ta,tb)\ge t^2(ab)^2\to\infty$ as $t\to\infty$. If $a=0$, then $b\ne0$ and $f(ta,tb)=t^2b^2\to\infty$ as $t\to\infty$. The case $b=0$ is similar. However, if $0<x\to\infty$ and $y=x^{-3/2}$, then $$f(x,y)=\frac1x+\Big(x^2+\frac1{x^3}\Big) \exp\Big(-x-\frac1{x^4}\Big)\to0.$$