$g(x)$ is convex. Show that $h(x) = \max(g(x), m)$ is convex

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.

• math.stackexchange.com/questions/147475/… Nov 4 '19 at 4:00
• 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. Nov 4 '19 at 4:19

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