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Let $f$ and $g$ be two convex functions. I’m supposed to give informations on $\sup(f,g)$ and $\inf(f,g)$ but I have no idea what that means.

Could anyone enlighten me please?

Nick
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Pablito
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    Well, presumably you are being asked to say whether they are or are not convex. What have you tried? – Arturo Magidin Sep 07 '19 at 22:36
  • Well I’ve never seen $sup(f,g)$ before I just assumed it was a number to be honest. – Pablito Sep 07 '19 at 22:41
  • It means the function whose value at $a$ is the larger of $f(a)$ and $g(a)$; $\inf(f,g)$ is the function whose value at $x$ is the smaller of the values $f(x)$ and $g(x)$. And btw, these are exactly the kind of questions you need to ask your instructor/whoever assigned this problem to you: if you don’t understand the notation, chances are lots of other people don’t either, and the instructor needs to know that so they can explain it to everyone. – Arturo Magidin Sep 07 '19 at 22:43
  • Oh ok It makes more sense now & I couldn’t ask them because it’s just a homework we were given and I didn’t notice the new notations ^^ – Pablito Sep 07 '19 at 22:46
  • hint: $\sup (f+g)(x)\le \sup f(x)+\sup g(x).$ – Matematleta Sep 07 '19 at 22:46
  • @Pablito: Most professors are perfectly fine receiving questions about homework at some time other than the instant it is given. Check with them on their protocol for contacting them with questions. It’s good practice to do so in any case, and to maintain communcation with the professor on what is difficult: you don’t want to give them a false impression that their homework is very easy because you are able to do it without having to ask anything about it (because you are getting help from hundreds of other people without their knowledge). – Arturo Magidin Sep 07 '19 at 22:50
  • Probably yes but my math teacher is bad so I’m not really into it haha. Anyway, I was trying to prove that $\forall (x,y) \in \mathbb{R}^2, \forall t \in [0,1], sup(f,g)(tx+(1-t)y) = max (f(tx + (1-t)y) , g(tx + (1-t)y)) \geq tmax(f(x),g(x)) + (1-t)max (f(y),g(y)) = t sup(f,g)(x) + (1-t) sup(f,g)(y)$ but I can’t quite find the right way to be honest. I always end up with < then > so I can’t conclude from this ^^ – Pablito Sep 07 '19 at 23:08
  • Sorry to say this then, but that’s absolutely the wrong attitude; you are headed for failure with that attitude. – Arturo Magidin Sep 08 '19 at 00:56

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Let $x,y\in\mathbb R$ and $t\in[0,1]$. We know that $$t\sup(f,g)(x)+(1-t)\sup(f,g)(y)=t\max(f(x),g(x))+(1-t)\max(f(y),g(y)).$$ But observe that the right-hand side is greater than or equal to $tf(x)+(1-t)f(y)\ge f(tx+(1-t)y)$. Similarly, it is greater than or equal to $g(tx+(1-t)y)$. Thus it is at least the maximum of those two expressions. In other words, we have that \begin{align*}t\sup(f,g)(x)+(1-t)\sup(f,g)(y)&\ge\max(f(tx+(1-t)y),g(tx+(1-t)y))\\&=\sup(f,g)(tx+(1-t)y)).\end{align*} Thus $\sup(f,g)$ is convex.

Regarding $\inf(f,g)$, we cannot say that is is either convex or concave. For instance, if $f(x)<g(x)$ for all $x$, clearly, $\inf(f,g)$ is convex. On the other hand, consider $f(x)=x^2$ and $g(x)=x^4$. Both are convex, but taking $x=0.9$, $y=1.1$, and $t=\frac12$ gives us that $$t\inf(f,g)(x)+(1-t)\inf(f,g)(y)=0.93305<1=\inf(f,g)(tx+(1-t)y).$$ So in this case $\inf(f,g)$ isn't convex. (In fact, it isn't concave either; just take $x,y\ge1$.)

boink
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