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Is there any algorithm, when given a vector $x$ of $n$ dimension, $k_1$($L^1$-norm) and $k_2$($L^2$-norm), find the vector $v$ of dimension $n$ having $\sum_i(|v_i|)=k_1$ and $\sum(v_i^2)=k_2$ which is closest to $x$.

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I don't understand. Do you mean $\sum |v_i|$ should be close to $\sum |x_i|$ and $\sum (v_i)^2$ should be close to $\sum (x_i)^2$? Then $x$ is a perfect choice. Or a vector $v$ with the same norms should be close to $x$ in some sense? I don't know which sense you mean, but $x$ still seems to be a perfect choice. –  savick01 Jan 9 '12 at 18:02
I mean the resultant vector $v$ whose L1 norm is $k_1$ and L2 norm is $k_2$. Given vector $x$ may not have those norm value. Find a vector $v$ which is close to $x$ (may be euclidean distance is minimum) –  Learner Jan 9 '12 at 18:37
I see. You can intersect $L^1$ $k_1$-sphere and $L^2$ $k_2$-sphere (both around zero) - if it is not empty, then it should be a manifold (or a sum of manifolds maybe), so you can use Lagrange's method (en.wikipedia.org/wiki/Lagrange_multiplier) to minimalise the value of euclidean distance on it. –  savick01 Jan 9 '12 at 18:56
Dear Learner, it would be very helpful if you corrected your title so that it does not have typos and is more explicit about what you want. –  Mariano Suárez-Alvarez Jan 9 '12 at 20:11
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