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Consider the function $f:D\rightarrow\mathbb{R}$ for $D\subset\mathbb{R}^n$ an open convex set. Furthermore, suppose that $g(t)=f(t\boldsymbol{x})$ is convex for all $\boldsymbol{x}\in D$. Is it true that $f$ is convex? Or is there a counter example?

I haven't been able to find a counter example.

Set
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2 Answers2

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The statement is trivially true for $n=1$, but false for all $n \ge 2$. As a counter-example, define $$ f: \mathbb R^n \to \mathbb R, \, f(x_1, x_2, \ldots, x_n) = | x_1\, x_2 | $$ For fixed $\boldsymbol{x} = (x_1, x_2, \ldots, x_n)$, $$ g: \mathbb R \to \mathbb R, \,g(t) = f(t \boldsymbol{x}) = t^2 | x_1\, x_2 | $$ is convex, but $f$ is not convex because $$ \frac 14 = f(\frac 12, \frac 12, 0, \ldots) \nleq \frac 12 f(1, 0, 0, \ldots) + \frac 12 f(0, 1, 0, \ldots) = 0 \quad. $$

Martin R
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It can be easily proved that a function $f$ is convex if and only if $t \mapsto f(x_0+th)$ is convex (in $t$) for any $x_0$ and $h$. This is stated here, and follows from the definitions. I don't believe that you can conclude, if you ignore translations, i.e. $x_0$.

Siminore
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  • I know you can't for a general domain $D$, but if we assume that $D$ is open and convex I was wondering if you could. – Set Jan 03 '15 at 11:33
  • Convexity is equivalent to convexity of the restriction to any line segment contained in $D$. You are considering only line segments that begin at the origin. I am skeptical... – Siminore Jan 03 '15 at 12:07
  • skeptical that it's true? – Set Jan 03 '15 at 12:20