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Given that $g(x)$ is a convex and monotonically nondecreasing function of a single variable, like, $g(x_{1}) \leq g(x_{2})$. And given that $f$ is a convex function on the convex set in $R^{n}$, how would I show that the composite function $g(f(x))$ is convex on $\Omega$?

I found online that $g(f(x))$ is convex if:

$(g(f(x)))''' = g''f(x))f'(x)^2 + g'(f(x))f''(x)$

But I am not sure how to show or prove this, as my original question states.

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Hint: Use the following definition for a convex function: $$h(x) \text{ convex } \Leftrightarrow\ h(tx_1+(1-t)x_2) \le th(tx_1) + (1-t) h(x_2).$$

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