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Let us assume $\bf{w}\in C^M$ is a complex vector with $M$ elements. Can we say anything about the convexity or concavity of function $f(\bf{w})=\log(1+\bf{w}^HR\bf{w})$? $R$ is a positive definite matrix and $R^H=R$.

Edit 1: $R$ is $M \times M$ and if we assume that $\bf{w}$$=[w_1,w_2,\ldots,w_M]^T$, then we also know that $\forall m: |w_m| \leq 1$.

Edit 2: Although $\bf{w}$ is a complex vector, $f$ is a real-valued function because we know that $R=\bf{h \times h^H}$ where $\bf{h}$ is a known vector. In other words, $f(\bf{w})$ $=\log(1+|\bf{w^H h}|^2)$.

Shahram Shahsavari
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  • You might need specify the region where you are interested to work with this function – Red shoes Aug 12 '17 at 18:48
  • the only restriction is that if we assume $\bf{w}$$=[w_1,w_2,\ldots,w_m]^T$, then $\forall m: |w_m| \leq 1$. – Shahram Shahsavari Aug 13 '17 at 01:35
  • Just to be 100% sure, what does "convexity" of $f(w)$ mean here? That for all $w,z\in\mathbb C^M$, and all real $t\in[0,1]$, you have $f(tz+(1-t)w)\le tf(z) + (1-t)f(w)$ ? Or instead, the subject discussed here ? – kimchi lover Aug 15 '17 at 00:08
  • good point! Although $\bf{w}$ is a complex vector, $f$ is a real valued function because we know that $R=\bf{h \times h^H}$ where $\bf{h}$ is a known vector. In other words, $f(\bf{w})$ $=\log(1+|\bf{w^H h}|^2)$ – Shahram Shahsavari Aug 15 '17 at 20:24

1 Answers1

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For at least one $R$ the answer has to be no. Because the function $x\mapsto \log(1+x^2)$ is neither convex or concave on $\mathbb R$.

In greater detail, added 15 August. The answer is always no, no matter what the positive definite Hermitian matrix $\mathbf R$ is. For a vector space $V$ over the reals, if $f:V\to\mathbb R$ is convex, its restriction $f_{|A}$ to any affine subspace $A\subset V$ is also convex; similarly if $f$ is concave so is $f_{|A}$. For the problem at hand, let $A$ be the one dimensional real subspace of $\mathbb C^M$ generated by $\mathbf a\in\mathbb C^M$. If $f$ is convex or concave so must the function $x\mapsto f_{|A}(\mathbf ax) = f(\mathbf ax) = \log(1+\mathbf{(a^HR a)} \,x^2)$. But this function is neither convex nor concave, so the original function $f$ is neither convex or concave.

kimchi lover
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