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Any constant function $g$ would do that, but that is useless for optimization problems.

If we add the condition that $g$ is not constant then the answer is no.

W.l.o.g. assume that $g(y_1) < g(y_2)$ for some $y_1 < y_2$. The convexity of $g$ implies that $g$ is strictly increasing for $y \ge y_2$. Now assume that $f \circ g$ is convex. Then for each closed interval $I \subset \Bbb R$, $f \circ g$ assumes its maximum at one of the boundary points.

Then for $g(y_2) < a < b$ $$ \max \{ f(x) \mid a \le x \le b \} = \max \{ (f\circ g)(y)) \mid g^{-1}(a) \le y \le g^{-1}(b) \} = \\ \max( (f\circ g)(g^{-1}(a)), (f\circ g)(g^{-1}(b)) ) = \max(f(a), f(b)) $$

and that is not satisfied for many functions, e.g. not for $f(x) = \sin(x)$.

Any constant function $g$ would do that, but that is useless for optimization problems.

If we add the condition that $g$ is not constant then the answer is no, even for functions defined on (finite or infinite) intervals. More precisely:

Let $I, J \subset \Bbb R$ be intervals, and $g: I \to J$ a non-constant convex function. Then there is a continuous function $f: J \to \Bbb R$ such that $f \circ g$ is not convex.

The idea is that a convex function assumes it maximum on any closed interval at one of the boundary points of the interval, and that this property is preserved under an (injective) change of variable.

Proof: W.l.o.g. assume that $g(x_0) < g(x_1)$ for some $x_0 < x_1$. The convexity of $g$ implies that $g$ is strictly increasing on any interval $[x_1, x_2] \subset I$. Then $g$ maps $[x_1, x_2]$ bijectively onto $[y_1, y_2] := [g(x_1), g(x_2)] \subset J$.

Now we can define $f: J \to \Bbb R$ as $$ f(y) = \sin\left( \pi \frac{y-y_1}{y_2-y_1}\right) \, . $$

If $f \circ g$ were convex then $$ \begin{aligned} \max \{ f(y) \mid y_1 \le y \le y_2 \} &= \max \{ (f\circ g)(x) \mid x_1 \le x \le x_2 \} \\ &= \max( (f\circ g)(x_1), (f\circ g)(x_2) ) \\ &= \max(f(y_1), f(y_2)) \\ & = 0 \end{aligned} $$

and that is not obviously not the case.

Any constant function $g$ would do that, but that is useless for optimization problems.

If we add the condition that $g$ is not constant then the answer is no, even for functions defined on (finite or infinite) intervals. More precisely:

Let $I, J \subset \Bbb R$ be intervals, and $g: I \to J$ a non-constant convex function. Then there is a function $f: J \to \Bbb R$ such that $f \circ g$ is not convex.

W.l.o.g. assume that $g(y_1) < g(y_2)$ for some $y_1 < y_2$. The convexity of $g$ implies that $g$ is strictly increasing for $y \ge y_2$.

If $f \circ g$ is convex, then for each closed interval in $I$, $f \circ g$ assumes its maximum at one of the boundary points.

Then for $g(y_2) < a < b \in I$ $$ \max \{ f(x) \mid a \le x \le b \} = \max \{ (f\circ g)(y)) \mid g^{-1}(a) \le y \le g^{-1}(b) \} = \\ \max( (f\circ g)(g^{-1}(a)), (f\circ g)(g^{-1}(b)) ) = \max(f(a), f(b)) $$

and that is not satisfied for $f(x) = \sin(kx)$ with $k$ sufficiently large.

Any constant function $g$ would do that, but that is useless for optimization problems.

If we add the condition that $g$ is not constant then the answer is no.

W.l.o.g. assume that $g(y_1) < g(y_2)$ for some $y_1 < y_2$. The convexity of $g$ implies that $g$ is strictly increasing for $y \ge y_2$. Now assume that $f \circ g$ is convex. Then for each closed interval $I \subset \Bbb R$, $f \circ g$ assumes its maximum at one of the boundary points.

Then for $g(y_2) < a < b$ $$ \max \{ f(x) \mid a \le x \le b \} = \max \{ (f\circ g)(y)) \mid g^{-1}(a) \le y \le g^{-1}(b) \} = \\ \max( (f\circ g)(g^{-1}(a)), (f\circ g)(g^{-1}(b)) ) = \max(f(a), f(b)) $$

and that is not satisfied for many functions, e.g. not for $f(x) = \sin(x)$.

Any constant function $g$ would do that, but that is useless for optimization problems.

If we add the condition that $g$ is not constant then the answer is no.

W.l.o.g. assume that $g(y_1) < g(y_2)$ for some $y_1 < y_2$. The convexity of $g$ implies that $g$ is strictly increasing for $y \ge y_2$. Now assume that $f \circ g$ is convex. Then for each closed interval $I \subset \Bbb R$, $f \circ g$ assumes its maximum at one of the boundary points.

Then for $g(y_2) < a < b$ $$ \max \{ f(x) \mid a \le x \le b \} = \max \{ (f\circ g)(y)) \mid g^{-1}(a) \le y \le g^{-1}(b) \} = \\ \max( (f\circ g)(g^{-1}(a)), (f\circ g)(g^{-1}(b)) ) = \max(f(a), f(b)) $$

and that is not satisfied for many functions, e.g. not for $f(x) = \sin(x)$.

Any constant function $g$ would do that, but that is useless for optimization problems.

If we add the condition that $g$ is not constant then the answer is no, even for functions defined on (finite or infinite) intervals. More precisely:

Let $I, J \subset \Bbb R$ be intervals, and $g: I \to J$ a non-constant convex function. Then there is a function $f: J \to \Bbb R$ such that $f \circ g$ is not convex.

W.l.o.g. assume that $g(y_1) < g(y_2)$ for some $y_1 < y_2$. The convexity of $g$ implies that $g$ is strictly increasing for $y \ge y_2$.

If $f \circ g$ is convex, then for each closed interval in $I$, $f \circ g$ assumes its maximum at one of the boundary points.

Then for $g(y_2) < a < b \in I$ $$ \max \{ f(x) \mid a \le x \le b \} = \max \{ (f\circ g)(y)) \mid g^{-1}(a) \le y \le g^{-1}(b) \} = \\ \max( (f\circ g)(g^{-1}(a)), (f\circ g)(g^{-1}(b)) ) = \max(f(a), f(b)) $$

and that is not satisfied for $f(x) = \sin(kx)$ with $k$ sufficiently large.