Intuition for when equality between norms is **not** obtained. Given two norms $\vert| \textbf{x} \vert|_p$ and $\vert|\textbf{x}\vert|_q$ where $1 \leq p \leq q < \infty$, it will hold that $\vert|\textbf{x} \vert|_q \leq  \vert|\textbf{x} \vert|_p$.
Is there an intuition or description of under what conditions equality is not met?
 A: I believe that there are only isolated cases when the two norms are equal, namely when $x=C e_i$ for some constant $C$ and where $e_i$ is the $i$-th basis vector in $\mathbb{R}^{n}$. Consider two dimensions. The unit ball for $p=1$ looks like a diamond, for $p=2$, it looks like a circle, and for $p=\infty$, it looks like a square. As $p$ increases, that ball gets interpolated at each step. The only points that remain common to each of those balls are $(1,0),(-1,0),(0,1),(0,-1)$. I've attached a photo of this. This would be the intuitive way to think about it.

To show this formally, consider the function
\begin{align}
f(x) = \|x\|_{p} - \|x\|_{q} \geq 0.
\end{align}
To find where it is equal to $0$ is to ask when they are equal. This is equivalent to finding where a minimum lies, so let's take partial derivatives. This means
\begin{align}
& \frac{\partial f}{\partial x_i} = 0 \\
& \Rightarrow |x_i| = \left( \frac{p}{q} \frac{\|x\|_{p}^{1-p}}{\|x\|_{q}^{1-q}} \right)^{\frac{1}{p-q}}
\end{align}
Now, if we constrain each $x_i$ to this equation, we see that $|x_i|=|x_j|$ for each $i,j$, so we may as well relabel this constant $|x_i| = C$. Then note,
\begin{align}
\|x\|_{p} = n^{\frac{1}{p}} C \neq n^{\frac{1}{q}} C = \|x\|_{q}
\end{align}
whenever $n > 1$. This means that the minimum on the interior of the $q$-unit ball (the compact domain I am restricting to since it also contains the $p$-unit ball) is $\textit{strictly}$ greater than zero. Therefore, any equality to $0$ must be on the boundary. Then they are certainly equal when $x = \pm e_{i}$ for unit vector $e_i$, but I would have to work out other parts of the boundary. The intuition from the $2$D case would tell you that it only works when $x = \pm e_{i}$ because the deformation of the unit ball works similarly in higher dimensions. Also, I may have made an algebraic mistake at some point; let me know, and I can edit.
