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I am trying to prove that the perspective of a convex function is convex.

Problem Statement
I'm trying to prove that $f$ is convex $$f(x,t) = tg(x/t)$$ where $g$ is a convex function, $x\in \mathbb{R}^n$, and $t>0$ is a positive scalar.

Proof
Since $g$ is convex, then $$g(\lambda x_1 + (1-\lambda)*y_1) \le \lambda g(x_1) + (1-\lambda)*t*g(y_1) \,\, \forall x_1, y_1$$ Let $x_1=x/t$ and let $y_1=y/t$, and multiply both sides by $t$: $$t*g(\lambda x/t + (1-\lambda)*y/t) \le \lambda t * g(x/t) + (1-\lambda)*t*g(y/t)$$ Using definition of $f$ observe: $$t*g(\lambda x/t + (1-\lambda)*y/t) = f(\lambda x + (1-\lambda)*y, t)$$ and $$ \lambda t * g(x/t) + (1-\lambda)*t*g(y/t) = \lambda f(x,t) + (1-\lambda)*f(y,t)$$ Substituting those into the inequality: $$f(\lambda x + (1-\lambda)*y, t) \le \lambda f(x,t) + (1-\lambda)*f(y,t)$$ which by definition means that $f$ is convex.

My proof doesn't use epigraphs, although almost every other proof I've seen does. E.g. this proof in number 6.

I want to know if my proof is acceptable without having used epigraphs? Thanks.

makansij
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    Your proof shows that, given any $t_0 > 0$, the single-variable map $x \mapsto f(x, t_0)$ is convex. You would need the convex combinations to extend to the second argument as well in order to establish convexity in full. But, there's no issue in general with using inequalities like this in lieu of epigraph arguments. – user804886 Jul 16 '20 at 03:57
  • Thanks @user804886. I understand your point about $t_0 > 0$, and I understand that "there's no issue in general...".

    But I don't understand why this only shows convexity for a single-variable map $x \mapsto f(x,t_0)$. Where in the argument do I make that assumption? Thanks.

     – makansij Jul 16 '20 at 14:59
  • Are you saying that this argument would be valid if there were a one-to-one mapping from $x/t$ to $x_1$? – makansij Jul 16 '20 at 15:06
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    In order to show convexity, you'd need to show that, given $x_1, x_2 \in \Bbb{R}^n$, $t_1, t_2 > 0$, and $\lambda \in [0, 1]$,$$f(\lambda x_1+(1-\lambda)x_2, \lambda t_1+(1-\lambda)t_2)\le\lambda f(x_1,t_1)+(1-\lambda)f(x_2,t_2).$$This shows convexity between any two points $(x_1, t_1), (x_2, t_2)$. You showed convexity between $(x, t)$ and $(y, t)$, i.e. two points with the same value of $t$. – user804886 Jul 16 '20 at 15:25
  • I see. So, in my proof above, in order to cover all possible combinations of $x_1$ and $y_1$, we need to consider that $t$ might be different for $x_1$ and $y_1$. Is that correct? I'll try to remediate and edit my proof. – makansij Jul 16 '20 at 17:47
  • To understand this at another level, I looked at another proof for which I thought it would be necessary to consider all possible combinations of two inputs: particulary in proving the convexity of sum of two functions $f$ and $g$: [here] (https://math.stackexchange.com/questions/325952/sum-of-strictly-convex-and-convex-functions). But to confirm with you, since there is only one input to $(f+g)$, we don't need to consider all possible combinations of inputs to $f$ and $g$? – makansij Jul 16 '20 at 18:04
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    To both your questions, yes. Convexity is a property of real-valued functions on vector spaces (or convex subsets of vector spaces), and the inequality satisfied has to hold for any convex combination of points in the vector space. In the above example, the subset of the vector space is $\Bbb{R}^n \times (0, \infty) \subseteq \Bbb{R}^{n+1}$. In the summation question, the only function we are proving convex is $f+g$ (using the fact that $f$ and $g$ are convex). Considering convex combinations there would be counterproductive, as $f$ and $g$ are not vectors in the domain of the function $f+g$. – user804886 Jul 17 '20 at 02:42
  • @user804886 Thanks for clearing that up I learned a lot from you! – makansij Jul 18 '20 at 23:11

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