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I am tasked to prove by the definition of convex sets that $S=\{\vec{x} : x_2 \geq e^{-x_1}\}$ is convex. It is simple to show this by considering the function $f(x) = e^{-x_2}-x_1$ and utilizing lower-level sets, but I would like to prove this directly by the definition of convex sets (i.e. the line segment between two points in $S$ is also in $S$).

So far, I've let $\vec{u}, \vec{v} \in S$, where $u=(u_1, u_2)$ and $v=(v_1, v_2)$ and established that our goal is to show that $\lambda u_2 + (1-\lambda)v_2 \geq e^{-(\lambda u_1 + (1-\lambda) v_1)}$.

From there, I've claimed that since $u_2 \geq e^{-u_1}$ and $v_2 \geq e^{-v_1}$, then $$\lambda u_2 + (1-\lambda)v_2 \geq \lambda e^{-u_1}+(1-\lambda)e^{-v_1}.$$ However, I don't know how to show that $\lambda e^{-u_1}+(1-\lambda)e^{-v_1} \geq e^{-(\lambda u_1 + (1-\lambda) v_2)}$ to conclude. So, how does one show that inequality?

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    $\begingroup$ A good start to a tasteful argument. Do you know if $x\mapsto \exp(-x)$ is convex? $\endgroup$ Sep 21 '18 at 2:13
  • $\begingroup$ Visually speaking, I would guess that graph($f$) is nonconvex but epi($f$) is convex, if $f(x) = e^{-x}$. $\endgroup$ Sep 21 '18 at 2:17
  • $\begingroup$ The epigraph of $\exp(x)$ and the epigraph of $\exp(-x)$ ought to be similar, I'd think: one is the mirror image of the other. $\endgroup$ Sep 21 '18 at 2:19
  • $\begingroup$ Is this a hint on showing that the inequality above is true? Or that I should consider the convex function $f(x) = e^{-x}$ and show that $S$ is a lower level set of a convex function? $\endgroup$ Sep 21 '18 at 2:24
  • $\begingroup$ Your $S$ is the epigraph of $f:x\mapsto\exp(-x)$. I don't know how your definitions are arranged, that is, what your precise basic definition of "convex" is. But at some stage you must know that the epigraph of a convex function is convex, and conversely. So. If you can prove $f$ is convex, you are done. $\endgroup$ Sep 21 '18 at 2:29
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The statement of your problem and discussion in comments has confused me a bit.

As I understand your question, you want to prove your 2-dimensional subset $S$ is convex by verifying that for any two points in $S$ the line segment connecting them is contained in $S$. Let $f:x\mapsto\exp(-x)$. You have reduced your problem to showing that for all $u,v\in S$, and $\lambda\in[0,1]$ you have $$ \lambda f(u_1)+(1−\lambda)f(v_1)≥f(\lambda u_1+(1−\lambda)v_1).$$ This is precisely the defining condition for $f$ to be a convex function of a single variable. So to complete your proof you need to show that $f$ is convex.

You seem to think that this argument is the same as showing that $S$ is a lower level set of the convex function $g(x,y)=\exp(-x)-y$ of two variables. Granted, they are close, but in my opinion they are not identical. To me the "$f$" proof is simpler than the "$g$" proof because I cannot see how you can possibly show $g$ is convex without also showing (or knowing) that $f$ is convex. The $f$ proof is, I think, more elegant, and contains more of the ideas about convexity you should keep in your head forever.

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