Given two row-vectors $u,v\in[-1,1]^n$, I would like to upper bound the Frobenius norm of the difference of their outer products. Formally, I would like an upper bound on $ \| uu^T - vv^T \|_F$. The upper bound should be a function of $\|u - v\|_2$, where $\|\cdot\|_2$ is the Euclidean norm. Do we know such a bound?
Since the entries of $u$ and $v$ are bounded, it is my hunch that an inequality of the form $ \| uu^T - vv^T \|_F \leq O(\sqrt{n}) \cdot \| u-v\|_2$ or $ \| uu^T - vv^T \|_F \leq O(1) \| u-v\|_2^2$ should hold.
What I tried so far:
I have tried using the definition of the Frobenius norm which gives that $ \| uu^T - vv^T \|_F^2 = \sum_{ij} (u_i u_j - v_i v_j)^2 = \sum_{ij} (u_i u_j)^2 - 2 u_i u_j v_i v_j + (v_i v_j)^2$. This is not very dissimilar from $\|u-v\|_2^4$ (after expanding) and I tried to make a few case distinctions based on the signs of $u_i u_j v_i v_j$ but this led nowhere (I spare you the details).
I also tried the brute-force upper bound $\sum_{ij}(u_i u_j - v_i v_j)^2 \leq n^2 \max_{ij}\{ (u_i u_j - v_i v_j)^2 \}$ and trying to relate this to $(u_i - v_i)^2$. However, this is tricky because $(u_i u_j - v_i v_j)^2 = (u_i u_j)^2 - 2 u_i u_j v_i v_j + (v_i v_j)^2$ but $(u_i - v_i)^2 = u_i^2 - 2 u_i v_j + v_i^2$; thus, even though I know that $(u_i u_j)^2 \leq u_i^2$ and $(v_i v_j)^2 \leq v_i^2$, I get problems with the term in the middle since if $u_i u_j v_i v_j$ is positive then we may be subtracting less than $u_i v_i$.