Converting trace to $2$-norm

Question

Assuming $x$ and $y$ are two vectors, I have an optimiztion problem with the following objective function

$$f(x,y)=\mbox{tr}((y-x)x^{\top})$$

Is it possible to convert the above to a norm-$2$, like $f(x,y)=\| \cdot \|_2^2$?

Answer 1

If $u,v$ are vectors then $\mathrm{trace}\left(uv^{\top}\right) = u\cdot v$, where $\cdot$ is the inner product.

So $$\mathrm{trace}\left((y-x)x^{\top}\right)=(y-x)\cdot x=y\cdot x - \|x\|^2$$

Answer 2

$\text{trace}(X X^\top) = \sum_{i,j} X_{ij}^2 = \|X\|_2^2$.

$$f(x,y) = \text{trace}((y-x)x^\top) = \left\|\frac{y}{2}\right\|_2^2 - \left\|x-\frac{y}{2}\right\|_2^2$$