Suppose we have two pure state $|\psi\rangle$ and $|\phi\rangle$.
I was wondering whether the statement: $\||\psi\rangle\!\langle\psi|- |\phi\rangle\!\langle\phi|\|_{\rm tr}$ is at most the Euclidean distance between $|\psi\rangle$ and $|\phi\rangle$. I know that under the qubit condition, the trace distance is exactly half of Euclidean distance.
If it is right, then what should the vector representation be corresponding to these two pure states?
You always have the opposite inequality: $\|A\|_1\ge \|A\|_2$.
You see it easily from the fact that $\|A\|_1$ is the sum of the singular values of $A$, while $\|A\|_2^2$ is the sum of the squares of the singular values. Thus $\|A\|_1^2 \ge \|A\|_2^2$, as the square of a sum is always larger than the sum of squares, and thus $\|A\|_1 \ge \|A\|_2$.
For the other direction, as pointed out in the other answer, you can still say something, as long as you introduce the dimension of the space, or better still, the rank. For any matrix $A$ you have $$\|A\|_2\le \|A\|_1 \le \sqrt{\operatorname{rank}(A)} \|A\|_2,$$ with the upper bound obtained via Holder's inequality as per the other answer. More generally, you have $$\|A\|_q \le \|A\|_p \le \operatorname{rank}(A)^{1/p-1/q}\|A\|_q,$$ for all $1\le p\le q \le\infty$ such that $1/p+1/q=1$. This is mentioned e.g. in chapter 1 of Watrous' book. A way to prove this is using the more general inequality for Schatten norms: $$\|ST\|_a\le \|S\|_b \|T\|_c, \qquad \frac1a=\frac1b+\frac1c, \quad a,b,c\in[1,\infty].$$ Applying this with $S=A$, $T=\Pi_{\operatorname{supp}(A)}$, $a=p$, $b=q$, we get $$\|A\|_p \le \|A\|_q \|\Pi_{\operatorname{supp}(A)}\|_{(1/p-1/q)^{-1}} = \|A\|_q \operatorname{rank}(A)^{1/p-1/q},$$ and we must have $1/p-1/q\ge1$, thus $q\ge p$.
In the special case of Hermitian matrices you can also use a simpler proof: given Hermitian $A$, let $U=\Pi_+-\Pi_-$, with $\Pi_\pm$ projections onto the spaces spanned by eigenvectors of $A$ with positive and negative eigenvalues, respectively. Then $\operatorname{tr}(A U)=\operatorname{tr}|A|$, and thus from Holder we get $$|\operatorname{tr}(AU)|=\operatorname{tr}|A| \le \|A\|_2 \|U\|_2 = \sqrt{\operatorname{rank}(A)} \|A\|_2,$$ because $\|U\|_2=\|\Pi_{\operatorname{supp}(A)}\|_2=\sqrt{\operatorname{rank}(A)}$ and $\operatorname{tr}|A|=\|A\|_1$. This is again a special case of a "tracial matrix Holder inequality", see this MO post.
Specialising to the case of pure states, everything's simple: you have $$\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1 = 2\sqrt{1-F}$$ with $F\equiv |\langle\psi|\phi\rangle|^2$, and I used the notation $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$. This follows from a direct calculation of the eigenvalues of $\mathbb{P}_\psi-\mathbb{P}_\phi$. Also $$\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_2^2 \equiv \operatorname{tr}[(\mathbb{P}_\psi-\mathbb{P}_\phi)^2] = \operatorname{tr}(\mathbb{P}_\psi+\mathbb{P}_\phi-2\mathbb{P}_\psi\mathbb{P}_\phi) = 2 (1-F).$$ Thus $$\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1 = \sqrt 2\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_2.$$ By contrast, the Euclidean distance between the ket vectors themselves reads $$\||\psi\rangle-|\phi\rangle\|_2^2 = 2 - 2\operatorname{Re}\langle\psi|\phi\rangle \ge 2(1-\sqrt F).$$ Furthermore $$2(1-\sqrt F) \ge \left(\frac{\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1}{2}\right)^2=1-F$$ always holds, because $$2(1-\sqrt F)-(1-F) = 1 - 2\sqrt F + F = (1-\sqrt F)^2 \ge0.$$ If follows that $$\||\psi\rangle-|\phi\rangle\|_2 \ge \frac{\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1}{2}.$$
Note that the "trace distance" is usually defined as half of $\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1$.
This can be inspected following Hölder's inequality with $p=q=2$:
$$||AI||_1\leq ||A||_2 ||I||_2$$ for any matrices $A$ and $I$ of dimension $d$. Setting $I$ to be the identity matrix, we find $$||A||_1\leq \sqrt{d}||A||_2$$ (also mentioned on Wikipedia). We thus do not have a proof that $||A||_1\leq ||A||_2$. For example, choosing any $d>1$ and $A=I$, we have $d=||I||_1 > ||I||_2=\sqrt{d}$.
Now that we know the trace norm is not always smaller than the Euclidean for an arbitrary matrix $A$, are there any special properties when $A$ is the difference between two states, $A=\rho-\sigma$?
I used Mathematica to check whether random density matrices have this special property and readily found examples that they don't: the 2-norm can be smaller than the 1-norm by more than a factor of 2
SeedRandom[dim];
r = With[{U = RandomVariate[CircularRealMatrixDistribution[dim]]},
U.(RandomReal[{0, 1}, dim] U\[Transpose])];
r = r/Tr[r];
s = With[{U = RandomVariate[CircularRealMatrixDistribution[dim]]},
U.(RandomReal[{0, 1}, dim] U\[Transpose])];
s = s/Tr[s];
A = r - s;
svals = SingularValueList[A];
n1 = Total[svals]
n2 = Sqrt[Total[svals^2]]
The outputs are n1=0.841057 and n2=0.295807. If you want to check the matrices yourself, they are $$\rho=\left( \begin{array}{cccccccccc} 0.0559484 & 0.0391899 & -0.00173254 & 0.0150553 & -0.00665287 & 0.0423966 & -0.0237883 & -0.00254245 & 0.00403863 & 0.0141285 \\ 0.0391899 & 0.114607 & -0.0355029 & 0.0449404 & 0.00791358 & 0.0208125 & -0.0166859 & 0.0136876 & -0.00365799 & 0.0577457 \\ -0.00173254 & -0.0355029 & 0.111272 & -0.00530915 & -0.0237596 & -0.00530644 & -0.0144068 & 0.025036 & 0.0104944 & -0.019394 \\ 0.0150553 & 0.0449404 & -0.00530915 & 0.0779406 & 0.0128229 & -0.00717888 & -0.0125049 & -0.00948783 & -0.0199239 & 0.0163034 \\ -0.00665287 & 0.00791358 & -0.0237596 & 0.0128229 & 0.0684224 & -0.00253051 & -0.00578879 & 0.00811702 & -0.00295013 & -0.0421486 \\ 0.0423966 & 0.0208125 & -0.00530644 & -0.00717888 & -0.00253051 & 0.166275 & 0.0078063 & -0.0168575 & 0.0515418 & -0.03558 \\ -0.0237883 & -0.0166859 & -0.0144068 & -0.0125049 & -0.00578879 & 0.0078063 & 0.106101 & -0.00781364 & 0.00546259 & 0.00478611 \\ -0.00254245 & 0.0136876 & 0.025036 & -0.00948783 & 0.00811702 & -0.0168575 & -0.00781364 & 0.078272 & -0.00208764 & 0.00160243 \\ 0.00403863 & -0.00365799 & 0.0104944 & -0.0199239 & -0.00295013 & 0.0515418 & 0.00546259 & -0.00208764 & 0.0919301 & -0.0107645 \\ 0.0141285 & 0.0577457 & -0.019394 & 0.0163034 & -0.0421486 & -0.03558 & 0.00478611 & 0.00160243 & -0.0107645 & 0.129231 \\ \end{array} \right)$$ and $$\left( \begin{array}{cccccccccc} 0.125917 & -0.0348865 & -0.0329077 & -0.00975323 & -0.00550824 & 0.0247898 & -0.00381375 & 0.0177037 & 0.0115744 & -0.00041872 \\ -0.0348865 & 0.0910228 & -0.00273958 & -0.0161907 & 0.010816 & -0.00247926 & 0.0103943 & 0.0055119 & 0.0220808 & 0.00719803 \\ -0.0329077 & -0.00273958 & 0.0671699 & 0.0178862 & -0.0175591 & 0.0186505 & -0.000334635 & -0.00606378 & 0.00894835 & -0.0522578 \\ -0.00975323 & -0.0161907 & 0.0178862 & 0.0976361 & 0.0494538 & 0.0108696 & -0.00845623 & 0.0206701 & -0.0305535 & 0.00555425 \\ -0.00550824 & 0.010816 & -0.0175591 & 0.0494538 & 0.122415 & 0.0255157 & 0.00702227 & -0.0282939 & 0.0271073 & 0.00524606 \\ 0.0247898 & -0.00247926 & 0.0186505 & 0.0108696 & 0.0255157 & 0.10571 & -0.00900667 & -0.0204656 & -0.00510038 & -0.0140047 \\ -0.00381375 & 0.0103943 & -0.000334635 & -0.00845623 & 0.00702227 & -0.00900667 & 0.0731207 & 0.00192831 & 0.0358366 & -0.00855586 \\ 0.0177037 & 0.0055119 & -0.00606378 & 0.0206701 & -0.0282939 & -0.0204656 & 0.00192831 & 0.0696558 & 0.0089128 & 0.0131745 \\ 0.0115744 & 0.0220808 & 0.00894835 & -0.0305535 & 0.0271073 & -0.00510038 & 0.0358366 & 0.0089128 & 0.13509 & -0.00930882 \\ -0.00041872 & 0.00719803 & -0.0522578 & 0.00555425 & 0.00524606 & -0.0140047 & -0.00855586 & 0.0131745 & -0.00930882 & 0.112262 \\ \end{array} \right).$$
The key point to realize here is that any norm distance of two pure states (in fact, any distance measure which is invariant under joint rotations $\lvert\phi\rangle \to U\lvert\phi\rangle$, $\lvert\psi\rangle \to U\lvert\psi\rangle$) can only depend on the angle $$\gamma = \langle \phi\vert\psi\rangle$$ between the two states, and can thus be studied by considering two qubits in states \begin{align} \lvert\hat\phi\rangle &= \vert 0\rangle\ , \\ \lvert\hat\psi\rangle &= \gamma\vert 0\rangle + \sqrt{1-|\gamma|^2}\lvert1\rangle\ . \end{align} Computing the values of the corresponding matrix norms is thus a $2\times 2$ matrix problem, which can be solved easily, and allows to relate any such norms by eliminating $\gamma$ from the equations.
