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The problem statement is:

Let $X \subseteq \mathbb{R}^n$ be closed and convex. Fixing $p \in \mathbb{R}^n$, is $p \cdot X = \{ y \in \mathbb{R} : y = p \cdot x \text{, for some } x \in X \}$ a closed set?

My intuition suggests that it would indeed be a closed set. If $X$ is assumed to be bounded, then it is continuous image of a compact set. The problem is I can't seem to find a rigorous way to explain what happens if $X$ is not bounded. Because $p$ is fixed and not allowed to depend on the input vector $x \in X$, I would think there would not be any problems like here. Any hints would be greatly appreciated.

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AWJ2108
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1 Answers1

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It may not be closed

$$X=\{(x,y)\in \Bbb R^2\,:\, xy\ge1\wedge x>0\}\\ p=(1,0)\\p\cdot X=(0,+\infty)$$

Geometrically: $X$ is the slice of half-plane that's over the hyperbola $y=\frac1x$ for positive $x$. Scalar product by $p$ yields the $x$-component, hence $p\cdot X$ is the domain of $f$.

  • Oh, right. I guess every example of X that I was thinking of was much nicer than that. Thanks! – AWJ2108 Feb 29 '16 at 02:26
  • One might add that the assertion holds, if $X$ is additionally assumed to be bounded (via a compactness argument). – gerw Feb 29 '16 at 07:27
  • @gerw The OP already knows that. However, in the related questions there is an interesting extension of "$X$ closed bounded$\Rightarrow$ $p\cdot X$ closed" in the case of Hilbert spaces. –  Feb 29 '16 at 12:19