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Let $g(x):=sup_{y\in R^n}f(x,y)$. Assume that for each fixed $y, f(x,y)$ is a convex function of x. Prove that $g(x)$ is a convex function.

I know that if ${\{f_i:i∈I\}}$ is a family of convex functions from a convex set $C $to $R$, then the same is true for $f=sup_{i∈I}f_i$. But how to prove it?

Walls
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  • https://math.stackexchange.com/questions/402919/pointwise-supremum-of-a-convex-function-collection?rq=1 , https://math.stackexchange.com/questions/410040/supremum-of-an-array-of-a-convex-functions?noredirect=1&lq=1 , https://math.stackexchange.com/questions/747588/why-is-pointwise-maximum-a-convex-function?noredirect=1&lq=1 ? – Calvin Khor Sep 21 '19 at 03:55
  • You are saying that you already know the answer. So what is the question? – Kavi Rama Murthy Sep 21 '19 at 05:33

1 Answers1

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Note that, for any $x$ and any $i$, $f_i(x) \le f(x)$. Hence, if $x, y$ are elements of the domain and $\lambda \in [0, 1]$, then for any index $i$, $$f_i(\lambda x + (1 - \lambda)y) \le \lambda f_i(x) + (1 - \lambda)f_i(y) \le \lambda f(x) + (1 - \lambda)f(y).$$ Consider the set of values of $f_i(\lambda x + (1 - \lambda)y)$, as $i$ ranges over the index set. We have just shown that this set bounded above by $\lambda f(x) + (1 - \lambda)f(y)$. However, by definition of $f$, its least upper bound is $f(\lambda x + (1 - \lambda)y)$. Therefore, $$f(\lambda x + (1 - \lambda)y) \le \lambda f(x) + (1 - \lambda)f(y),$$ proving convexity.

Theo Bendit
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