Showing that $\{\vec{x} : x_2 \geq e^{-x_1}\}$ is a convex set

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?

• A good start to a tasteful argument. Do you know if $x\mapsto \exp(-x)$ is convex? Sep 21 '18 at 2:13
• Visually speaking, I would guess that graph($f$) is nonconvex but epi($f$) is convex, if $f(x) = e^{-x}$. Sep 21 '18 at 2:17
• 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. Sep 21 '18 at 2:19
• 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? Sep 21 '18 at 2:24
• 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. Sep 21 '18 at 2:29

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.