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Although complexity analysis can give you some insight on the difficulty of your problem, it is unlikely that will settle your question in full-generality. For example: in the oracle model, a strongl …

Regarding difficulty results (i.e., lower bounds) for iterative methods, worst-case complexity results in the oracle model hold regardless of whether you provide the full subgradient to the algorithm. …

It took me some time, but I hope it still helps. Coming back to your problem I realized that the norm constraint is not convex (you should have the opposite inequality for convexity). Assuming you ha …

The most general form of such algorithms are named Mirror-Descent. This algorithm is an extension of gradient descent for non-Euclidean geometries. For a formal explanation on how multiplicative weig …

If you are interested in complexity results, I advice you to look at this paper https://arxiv.org/abs/1501.07242 by Belloni, Liang, Narayanan and Rakhlin. I think their results are nearly optimal when …

If you replace $y$ as you suggest, then your objective (only in terms of $x$) is given by a strongly convex function (the quadratic) plus a convex function. Since this defines a strongly convex object …

I am not aware of this successive projection algorithm: Can you provide a reference for it? In principle, I don't see a problem with what you are doing, because you only have $n$ nonnegativity constra …

There is a meaningful oracle model where you can obtain a provably optimal method when searching for approximate solutions: this is the "matrix-vector multiplication oracle", where you want to solve a …

I believe I can answer your question for the case where $E=\Delta_n:=\{x\geq 0:\,\,\sum_i x_i = 1\}$ is the standard $n$-dimensional simplex. My first thought when I saw your question was to consider …

Let $B = xx^T$, and $Y=(D+T)$, then $x^TY^{-1}x=\mbox{Tr}[Y^{-1}B]=\mbox{Tr}[\sqrt{B}^T Y^{-1} \sqrt{B}]$, where $B=\sqrt{B}\sqrt{B}^T$ is the Cholesky factorization. Note that $$\min \mbox{Tr}[\sqrt{ …

Your objective is convex, and can be written in conic form (to be precise, a mixed SDP and SOCP). By cyclicity of the trace, and since $X$ is Hermitian $$ Tr(AX^2)=Tr(XAX^H)=Tr(XF(XF)^H) $$ Now, the s …

In the closed convex case there are some fairly efficient algorithms, as long as you can efficiently project any point $x$ onto $A$ and $B$. This class of algorithms (named alternating projections, an …

The situation were your objective $f(x,y)=x^T A y$ is bilinear corresponds to the case of zero-sum games (http://www.inf.ed.ac.uk/teaching/courses/agta/lec4.pdf). If you remove the convexity assumptio …

Note that $f(t)=g(A(t))$, where $A(t_1,t_2)=A_0+t_1A_1+t_2A_2$ is affine and $g(B)=\lambda_{min}(B)$ which is concave on $B$. Now, to obtain the supergradient of this function you just need to use th …