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Let $g(x)$ be a convex function and let $m \in R$. Show that the function $h(x)$ defined by $h(x) = \max(g(x),m)$ is convex.

The definition of convexity is that $f(x)$ is convex for $x_1, x_2 \in R$ and $\lambda_1, \lambda_2 > 0$ s.t. $\lambda_1 + \lambda_2 = 1$ iff $$f(\lambda_1x_1 + \lambda_2x_2) \le \lambda_1f(x_1) + \lambda_2f(x_2)$$

So I know that if $h(x)$ is to be convex then $$h(\lambda_1x_1 + \lambda_2x_2) \le \lambda_1h(x_1) + \lambda_2h(x_2)$$ $$\max(g(\lambda_1x_1 + \lambda_2x_2),m) \le \lambda_1\max(g(x_1),m) + \lambda_2\max(g(x_2),m)$$

But this is where I am stuck. Should I break the proof into different cases based on different values of $m$? I feel like I'm very close to this proof but can't see the key step.

Thanks in advance for your help.

TenaliRaman
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meff11
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  • https://math.stackexchange.com/questions/147475/proving-that-the-maximum-of-two-convex-functions-is-also-convex – TenaliRaman Nov 04 '19 at 04:00
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    The intersection of two convex sets is convex. A half plane is convex and the epigraph of $f(x)$ is convex. Now show that $\max(f(x),m)$ is the intersection of a certain half plane and the epigraph and you're done. – CyclotomicField Nov 04 '19 at 04:19

2 Answers2

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$f(\lambda_1 x+(1-\lambda_2)y) \leq \lambda_1f(x)+(1-\lambda_2)f(y)\leq \lambda_1h(x)+(1-\lambda_2)h(y)$ because $f(x)\leq h(x)$ and $f(y)\leq h(y)$.

Also $m = \lambda_1 m+(1-\lambda_2)m \leq \lambda_1h(x)+(1-\lambda_2)h(y)$.

If two numbers are both $\leq \lambda_1h(x)+(1-\lambda_2)h(y)$ then their maximum is also $\leq \lambda_1h(x)+(1-\lambda_2)h(y)$.

Hence $h(\lambda_1 x+(1-\lambda_2)y) \leq \lambda_1h(x)+(1-\lambda_2)h(y)$.

Kavi Rama Murthy
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Intuitively i'll say that a convex function has at most one local minima. If $g(x)$ is all time increasing or decreasing. Your case is easy as $h(x)$ is a straight line with a part of a convex function above that line. If $m$ is that minima we are done, if $m$ is after or before that minima $h(x)$ is a line surrounded upperely by convex functions. On a graph $h(x)$ is convex.

Toni Mhax
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