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Let $f:\mathbb R^n \rightarrow [-\infty,+\infty]$. show that the perspectie of $f$ defined via

$$ g(x,y)=yf(x/y),\qquad x\in \mathbb R^n,y\in \mathbb R_+ $$ is convex iff $f(x)$ is a convex function.

$$\nabla g(x,y)=\left(\frac{\partial yf(x/y) }{\partial x}, \frac{\partial yf(x/y) }{\partial y}\right)= \left(\frac{y \partial f(x/y) }{\partial x}, \frac{\partial yf(x/y) }{\partial y}\right)\\= \left(y \frac{\partial f(x/y) }{\partial x} \frac{1}{y}, f(\frac{x}{y}) + \frac{\partial f(x/y) }{\partial y} (-\frac{x}{y})\right)$$

Ok. I have shown the $1$-st derivative. But how do I know it is convex? I do not know how to continue for the second derivative? $2$-nd derivative must be $> 0$.

Arian
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jester
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    normally you need also differentiability of $f$ for the derivative test. however in general you can always use the basic definition for convexity, that every segment line joining any of two points on the graph of $f$ does not lie below the graph. – Arian Mar 30 '18 at 10:00

1 Answers1

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Stephen Boyd's book proves this in section 3.2.6 using epigraphs. Exercise 3.33 hints that you can also do this directly from the definition of convexity.

Casey
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