Claim. Fix $0<k<\ell$ and an arbitrary constant $C$. If $n$ is sufficiently large, there does not exist
a function $f:\{0,\dots,n\}^k\to [0,Cn^{k/\ell}]^\ell$ such that $|f(x)-f(y)|\ge|x-y|^{k/\ell}$ for any $x,y\in\{0,\dots,n\}^k\subset\mathbb{R}^k$.
Proof. Assume by contradiction that such an $f$ exists. Given a closed cube $Q\subset\mathbb{R}^\ell$, we call $\delta(Q):=\frac{\# f^{-1}(Q)}{|Q|}$ its density (here $|Q|$ is the volume of the cube).
Lemma 1. Fix any $\epsilon>0$. There exist $N>0$ (integer) and $a_0>0$ (both depending only on $\epsilon$, $k$ and $\ell$) with the following property:
if a cube $Q$ has $\delta(Q)\ge\epsilon$ and sidelength $a\ge a_0$, then there exists a subcube $Q'\subset Q$ with sidelength $\frac{a}{N}$ and
$\delta(Q')\ge(1-N^{-\ell})^{-1}\delta(Q)$.
Proof. Split $Q$ into $N^\ell$ subcubes $\{Q_i\}$ (with sidelength $\frac{a}{N}$ and disjoint interiors). It suffices to prove that one of these cubes, say $Q_{i_0}$, has empty preimage: then it follows that
$$ a^\ell\delta(Q)=\# f^{-1}(Q)\le\sum_{i\neq i_0}\# f^{-1}(Q_i)\le(N^\ell-1)\left(\frac{a}{N}\right)^\ell\max_{i\neq i_0}\delta(Q_i), $$
so we can choose any $Q_i$ realizing the maximum as our $Q'$.
Assume by contradiction that all $Q_i$ intersect the image of $f$. Choose $x,y\in f^{-1}(Q)$ such that $|x-y|$ has the maximum possible value.
We can find a chain of at most $\ell N$ cubes $Q_{i_0},\dots,Q_{i_s}$ such that two consecutive cubes in the chain share an $(\ell-1)$-dimensional face
and such that $x\in Q_{i_0}$, $y\in Q_{i_s}$.
By hypothesis, for all $j=1,\dots,s-1$ we can find $x_j\in f^{-1}(Q_{i_j})$. We also put $x_0:=x$ and $x_s:=y$. We have $|f(x_j)-f(x_{j-1})|\le 2\ell\frac{a}{N}$,
so $|x_j-x_{j-1}|\le (2\ell)^{\ell/k}\left(\frac{a}{N}\right)^{\ell/k}$.
By the triangle inequality, we deduce
$$ |x-y|\le (\ell N)(2\ell)^{\ell/k}\left(\frac{a}{N}\right)^{\ell/k}=\ell(2\ell)^{\ell/k} a^{\ell/k}N^{1-\ell/k}. $$
Since all points in $f^{-1}(Q)$ have distance at most $|x-y|$ from $x$, we deduce that
$$ \epsilon\le\delta(Q)=a^{-\ell}\# f^{-1}(Q)\le a^{-\ell}(2|x-y|+1)^k\le a^{-\ell}\cdot 2^{k-1}(2^k|x-y|^k+1)\le 2^{2k-1}\ell^k(2\ell)^\ell N^{k-\ell}+2^{k-1}a^{-\ell}. $$
Since $k-\ell<0$, we obtain a contradiction if $N$ and $a_0$ are large enough. $\blacksquare$
Lemma 2. If $Q$ has sidelength in $\left[\frac{a_0}{N},a_0\right]$, then $\delta(Q)\le\delta_0$ (for some $\delta_0$ depending only on $N,a_0,k,\ell$).
Proof. For any distinct $x,y\in f^{-1}(Q)$ we have $|x-y|\ge 1$, so that $|f(x)-f(y)|\ge 1$. Thus, the cubes centered at $f(x)$ and $f(y)$ with sidelength $\frac{1}{\sqrt{n}}$ are disjoint. So the cube $\tilde Q$ with the same center as $Q$ and sidelength $a_0+\frac{1}{\sqrt{n}}$ contains $\# f^{-1}(Q)$ such disjoint cubes. Comparing the volumes, this gives
$$ \# f^{-1}(Q)\cdot\left(\frac{1}{\sqrt{n}}\right)^\ell\le\left(a_0+\frac{1}{\sqrt{n}}\right)^\ell, $$
so that $\delta(Q)\le\left(\frac{a_0}{N}\right)^{-\ell}\# f^{-1}(Q)\le N^\ell a_0^{-\ell}(a_0\sqrt{n}+1)^\ell$. $\blacksquare$
Proof of the claim (continued). Now the cube $Q_0:=[0,Cn^{k/\ell}]^\ell$ has density $\frac{(n+1)^k}{C^\ell n^k}\ge C^{-\ell}=:\epsilon$. Let $N$ and $a_0$ be the corresponding constants given by Lemma 1. Let then $\delta_0$ be the constant given by Lemma 2. Fix a positive integer $m$ such that
$$ \epsilon\cdot(1-N^{-\ell})^{-m}>\delta_0. $$
Assume $n$ is so large that
$Cn^{k/\ell}\ge N^m a_0$. Applying Lemma 1 to $Q_0$ we find a subcube $Q_1$ with density at least $\epsilon\cdot(1-N^{-\ell})^{-1}$.
Then we can apply Lemma 1 again to $Q_1$, and so on. We apply Lemma 1 as long as the sidelength of the cube is at least $a_0$, obtaining a finite sequence of cubes $Q_0,\dots,Q_s$, where $Q_s$ has sidelength $N^{-s}Cn^{k/\ell}\in\left[\frac{a_0}{N},a_0\right)$.
From $N^m a_0\le Cn^{k/\ell}\le N^s a_0$ we deduce $s\ge m$, so $\delta(Q_s)\ge(1-N^{-\ell})^{-s}\epsilon>\delta_0$. However, the cube $Q_s$ contradicts Lemma 2. $\blacksquare$
Theorem. If $0<k<\ell$, there does not exist a function $f:[0,1]^k\to\mathbb{R}^\ell$ such that $|f(x)-f(y)|\ge|x-y|^{k/\ell}$ for any $x,y\in [0,1]^k$.
Proof. Assume such an $f$ exists. Let $S_i:=\{x\in [0,1]^k:|f(x)|\le i\}$. The increasing sequence $\overline{S}_1,\overline{S}_2,\dots$
of closed sets covers $[0,1]^k$. By Baire's theorem, for some $i>0$ the set $\overline{S}_i$ contains a subcube $[a,a+b]^k$.
For any $x\in [a,a+b]^k\setminus S_i$ we choose a sequence $(x_m)\subseteq S_k$ with $x_m\to x$ and (up to extracting subsequences) we assume that $f(x_m)$ has a limit (this can be done since $|f(x_m)|\le i$). We replace $f(x)$ with this limit. The new function, still denoted by $f$, satisfies
$|f(x)|\le i$ and $|f(x)-f(y)|\ge|x-y|^{k/\ell}$ for any $x,y\in[a,a+b]^k$.
Now let us consider the function $g:\{0,\dots,n\}^k\to\mathbb{R}^\ell$ given by
$$ g(x):=\left(\frac{n}{b}\right)^{k/\ell}f\left(a+\frac{b}{n}x\right)+\left(\frac{n}{b}\right)^{k/\ell}(i,\dots,i), $$
which takes values into $[0,2\ell i b^{-k/\ell} n^{k/\ell}]$. For $n$ large enough, this $f$ must contradict the Claim. $\blacksquare$
