Consider the following proposition:
Proposition 1:
Let $\Omega$ be an open, convex subset of $\mathbb R^2$ and let $f:\Omega \to \mathbb R$ be a $C^1$ function with bounded gradient. Then $ \frac{f(x)}{1+|x|} $ is bounded.
Proof of proposition 1:
A $C^1$ function with bounded gradient over a convex domain is Lipschitz
(by the Mean Value Theorem). Hence, $\forall x,y\in \Omega,$
$$|f(x)-f(y)|\leq M |x-y|,$$ where $M>0$ is the Lipschitz constant.
If we fix an $y_0\in \Omega$ we have, $\forall x \in\Omega$
$$|f(x)|\leq |f(y_0)|+|f(x)-f(y_0)|\leq |f(y_0)|+ M|x-y_0|\leq |f(y_0)|+ M|x|+M|y_0|\leq C(1+|x|),$$ where $C=max\{|f(y_0)|+M|y_0|;M\}.$
Now consider another similar proposition:
Proposition 2:
Let $\Omega$ be an open, convex subset of $\mathbb R^2$ and let $f:\Omega \to \mathbb R$ be an uniformly continuous function. Then $ \frac{f(x)}{1+|x|} $ is bounded.
Proof of proposition 2:
$f$ is an uniformly continuous function over a convex domain. Hence, $\forall \epsilon>0\; \exists K>0 $ such that
$$|f(x)-f(y)|\leq K|x-y|+\epsilon\quad\quad\quad \forall x,y\in \Omega.$$
(See http://orfe.princeton.edu/~rvdb/tex/unif_cont/uc3.pdf for a proof of this fact.) Reasoning as in the proof of proposition 1, we can conclude that $|f(x)|\leq C(1+|x|).$
Note also that the two hypotesis "uniformly continuous" and "$C^1$ with bounded gradient" are, in general, independent. If the domain is convex the bounded gradient implies that the function is Lipschitz and hence also uniformly continuous. In general, however, it is possible to find uniformly continuous functions with unbounded derivative ($\sqrt x$ over $(0, +\infty)$) or functions with bounded gradient but not uniformly continuous ($f(x,y)$= "argument of $(x,y)$" over the annulus with a radius removed).
My questions are:
Is it possible to find a non-convex domain $\Omega$ and a function $f:\Omega\to \mathbb R$ $C^1$ with bounded gradient such that $ \frac{f(x)}{1+|x|} $ is unbounded?
Is it possible to find a non-convex domain $\Omega$ and a function $f:\Omega\to \mathbb R$ uniformly continuous such that $ \frac{f(x)}{1+|x|} $ is unbounded?
Is it possible to find a non-convex domain $\Omega$ and a function $f:\Omega\to \mathbb R$ uniformly continuous, $C^1$ with bounded gradient such that $ \frac{f(x)}{1+|x|} $ is unbounded?
Answer
I think the answer to $3$ is yes. Define
$$A_h= (0,1)\times (0,h)\cup (0,3)\times (h-1,h) \cup (2,3)\times (0,h.)$$
Think of $h> 0$ as large; actually $h>6$ should be enough to make the argument work. Then $A_h$ has the shape of an arch that is long in the vertical direction. (Good to draw a picture!)
We want to create a certain function $f$ in $A_h$ that has bounded gradient and is uniformly continuous. To do this, choose any smooth increasing $g$ on $\mathbb R$ such that $g=0$ on $(-\infty,1/3],$ $g=1$ on $[2/3,\infty).$ Put $M = \sup_{\mathbb R} g'.$ Now define, for $(x,y) \in A_h,$
$$\tag 1 f(x,y) = \begin{cases} hg(y/h) & (x,y) \in (0,1)\times (0,h)\\ h & (x,y) \in (0,3)\times (h-1,h)\\ h+hg(1-y/h) & (x,y) \in (2,3)
\end{cases}$$
Note that as we move clockwise in $A_h,$ $f$ grows from $0$ at the bottom left to $2h$ on the bottom right. This is useful, because $h$ is as large as we like while $A_h$ measures only $3$ from left to right!
We have $f$ smooth in $A_h$ and $|\nabla f| \le M$ there. We also have $|f(p)-f(q)| \le M |p-q|$ for all $p,q\in A_h$ with $|p-q| < 1$ (this is where $h>6$ comes in handy). Note that these estimates do not depend on $h;$ this is important. Also note the subregions of $A_h$ where $f$ is constant.
So the idea is to consider a sequence of disjoint arches like $A_h$ along the $x$-axis, at distance $1$ from each other, with varying heights $h,$ with corresponding functions as in $(1).$ We patch them together with an infinite strip of height $1$ along the bottom. In doing this we can have virtually any rate of growth, certainly enough to make $f(x)/|x|$ blast off to $\infty.$ We will have the gradient of the total function bounded by $M,$ and we will also have $|f(p)-f(q)| \le M |p-q|$ when $|p-q| < 1,$ giving the desired uniform continuity.
I'll stop here for now. Ask questions if you like.
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