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I'm not sure I properly understand how the below question is being posed. This is a question asked following lectures on linear operators and concavity, for context. To repeat, not looking for an answer, just help in understanding what's being asked of me.

$f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ is concave iff $\lambda \mapsto f(\lambda x + (1 - \lambda)y)$ is a concave map on $[0, 1]$ for every $x, y \in \mathbb{R}^{n}$.

To me this seems to apply something like a linear mapping $f(\lambda) = \sum_{i=1}^{n}(\lambda x + (1 - \lambda)y)$, but that doesn't feel right. Can anyone help me understand $f$, $\lambda$ and their relationship?

hillard28
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2 Answers2

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If you think of $x$ and $y$ as points in space, then the line segment connecting those two points is parameterized by the function $z(\lambda) := \lambda x + (1-\lambda) y$ for $\lambda$ in $[0,1]$. For example, $z(0)=y$, $z(1)=x$, and $z(1/2)$ is the midpoint of the line segment. The map $\lambda \mapsto f(z(\lambda))$ is a one-dimensional function; it's the restriction of $f$ to a particular line segment. The definition is just saying that $f$ is an n-dimensional concave function iff it is a 1-dimensional concave function when restricted to any line segment.

p.s.
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I assume you know the definition of concave map for a function $g: \mathbb{R} \rightarrow \mathbb{R}$.

The question Is asking you to consider the function $g(\lambda)=f(\lambda x+(1-\lambda) y)$ where $x,y$ are arbitrary (but fixed) vectors of $\mathbb{R}^n$ i.e. by the usual definition of concave Maps you can rewrite the question as:

Show that $f$ Is concave iif for any $\alpha \in [0,1]$ $$ g(\alpha \lambda_1+ (1-\alpha) \lambda_2) \ge \alpha g(\lambda_1)+ (1-\alpha) g(\lambda_2) $$

Marco
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