The infimum of a gradient over the whole $\mathbb{R}^d$ Let $\{f_k\}:\mathbb{R}^d\to\mathbb{R}$ be a sequence of $C^1$ functions which converges pointwise to 0. Is it true that
$$\lim_{k\to+\infty}\inf_{x\in\mathbb{R}^d}|\nabla f_k(x)|=0?$$
If $d=1$ I believe the result is true, but what happens if $d>1$? I believe counterexamples can be constructed, maybe building a sequence of functions for which the modulus of the gradient remains bounded away from 0 but the modulus of each partial derivatives oscillates between 0 and a certain value, keeping the modulus of the whole gradient above the treshold.
I shall present the proof for the case $d=1$.
Take a compact set $A\subset \mathbb{R}$ which has non empty interior, then clearly
$$\limsup_{k\to+\infty}\inf_{x\in\mathbb{R}^d}|f^{\prime}_k(x)|\leq\limsup_{k\to+\infty}\inf_{x\in A}|f^{\prime}_k(x)|.$$
Let us focus on the right hand side of the previous expression: suppose that there exists $l>0$ such that
$$\limsup_{k\to+\infty}\inf_{x\in A}|f^{\prime}_k(x)|\geq l.$$
Then $\forall k$, due to compactness, there exists $x_k\in A$ such that
$$\limsup_{k\to+\infty}|f^{\prime}_k(x_k)|\geq l.$$
But then, being $x_k$ a point of minimum, the following holds
$$\limsup_{k\to+\infty}|f_k^{\prime}(x)|\geq l\qquad\forall x\in A.$$
Passing to a subsequence that I won't relabel, this implies that $\forall k\in\mathbb{N}$, either $f_k^{\prime}(x)\geq l/2$ for every $x\in A$, or
$f_k^{\prime}(x)\leq -l/2$ for every $x\in A$ due to the continuity of $f_k^{\prime}$.
Now take an interval $[x,y]\in A$ (with $x<y$) and use the fundamental theorem of calculus to write
$$f_{k}(y)-f_k(x) = \int_x^yf_k^{\prime}(t)\textrm{d}t.$$
Taking the limit on both sides and using the pointwise convergence of the sequence $f_k$ leads now to a contradiction.
Maybe my proof is overly complicated but it seems to me that it works. Let me know what you think, especially of the case $d>1$.
 A: Here is the outline of the construction.
We shall start with considering a vertical half-strip going up, the left boundary of which will be called positive (shown in red) and the right boundary will be called negative (shown in blue) and carry out the following construction (generation 1 and generation 3 shown on the two next pictures; the rest of the strip work will be illustrated on generations 2):

This was generation 1.

And that is generation 3. How to proceed should be clear now (if not, look at the generation 2 below).
Notice that at generation $k$ we turn each side into a $3^{-k}$ dense tree and we have a winding strip between these trees one side of which is positive and the other one negative. Now expand each tree a bit into a domain with smooth boundary (the expansion can be made as small as we wish).

Our function will be $\varepsilon=3^{-k}$ on the boundary of the red domain and $-\varepsilon=-3^{-k}$ on the boundary of the blue domain.
We can make a smooth foliation of the winding yellow strip with transversal lines of length about $3^{-k}$ as shown (the parts where the boundaries are true parallel straight lines are trivial). We can make a descent along each line with gradient about $1$ most of the time while near the endpoints we can declare any normal derivative larger than $1$ we want as long as it smoothly depends on the point (the exact value will be determined by what happens inside the domains).
Inside the domains we will draw the function level lines as follows (effectively removing one generation growth a time):

The point is that we can move the level lines at speed $\le 1$ to keep the gradient above $1$, so we are forced to add $3^{-\ell}$ to the function when collapsing the level $\ell$ sprouts. Fortunately, the sum of the geometric progression is its largest term, so we keep the function at the level $3^{-\ell}$ around the part of the tree we added when going from generation $\ell-1$ to generation $\ell$. I'll call this part the $\ell$-th generation intervals. Note that they all are located on a fixed discrete grid of lines (I was too lazy to make that picture but I'll do it if somebody really wants it).
Generally the function increases in the red domain and decreases in the blue one according to the arrows below:

Now comes some analysis. This function is of order $3^{-k}$ outside the blue and red domains that can be made of any width $\delta_k>0$ we want. However, if we keep the picture in place, the original trees will be bad. So we will shift it by some decreasing sequence $\mu_k>0$ in both vertical and horizontal directions demanding that the intervals $(\mu_k-4\delta_k,\mu_k+4\delta_k)$ are disjoint. Then for each level $\ell$, the $\delta_k$ neighborhoods of the, say, horizontal lines on which the $\ell$-th generation intervals lie are eventually disjoint (that happens as soon as $\mu_k$ gets too small to make it possible for a shift of a grid line to intersect another shift of a different grid line). Thus, for any fixed $\ell\ge 0$, each fixed point can lie only in finitely many $\delta_k$ neighborhoods of the shifted $\le \ell$-th generation intervals, i.e., eventually the function drops below $3^{-\ell}$ and stays there.
This gives you an example in a half-strip. What was crucial was that we can make these small shifts and that the boundary lines have the starting point. The possibility to shift is due to the foliation of the plane into vertical lines and the transversal foliation into the horizontal ones. So we need to make something like that but covering everything. The solution is the foliation into double spirals and the transversal foliation into rays from the origin:

The spirals are unit width ones outside the disk of radius $1$ and logarithmic inside. This creates a nice coordinate system between the red and blue lines outside an arbitrarily small neighborhood of the origin and we can just replant our entire construction there. What to do near the origin? It turns out that we need to do nothing because the spirals themselves are already very close there (we need to gradually reduce the neighborhoods with $k$ to $0$, of course). So, if we stop the spirals at some point, extend them to narrow smooth domains and do the transversal foliation of the strip in between, we can just do the straight descent from red to blue as before without forcing any small values of the gradient:

After just a turn or two, our coordinate system kicks in and the further expansion of the strip is of no concern.
The result is that every point except the origin is eventually covered by the coordinate system and our tree nonsense and also decides between which two rotations of the red/blue spirals it stays (so the coordinates get eventually consistent), so it is good in terms of convergence to $0$. But the origin is always good (if you look at out functions closely, you'll realize that they are actually odd). The escapes along the spirals are, of course, the same as along the boundary lines of the half-strip.
That is all (modulo minor boring formalities related to $C^1$-smoothness).
Describing pictures in words is not my strong side, but if you have trouble understanding anything, feel free to ask questions.
