# Existence of Continuous Function

The following question requires the use of an easy theorem of calculus, but I am failing to see which one.

Let $g$ be non-constant and $C^{1}$ on some interval $I$. Show that for some subinterval $J \subseteq I$, there exists a continuous function $f$ such that the differential equation $y' = f(y)$ has the solution $y = g(x)$ on $J$.

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Don't you mean $y = g(x)$? – Javier Jan 15 '13 at 7:28
Sorry! Mistake in notation! Corrected. – user44069 Jan 15 '13 at 7:30
Isn't this just the fundamental theorem of calculus, in that differentiation and integration are loosely "inverse" operations? Since $g$ is $C^1$ we can always try $f(x)=g'(x)$ and $f$ will be continuous. – Gyu Eun Lee Jan 15 '13 at 7:33
@Stefan: Did you actually mean ‘$y'(x) = f(x)$’ instead of ‘$y'(x) = f(y(x))$’? – Haskell Curry Jan 15 '13 at 8:25
I am not really sure. I typed the problem as I have it in front of me. – user44069 Jan 15 '13 at 8:30

• Suppose that $g: I \to \mathbb{R}$ is non-constant and $g \in {C^{1}}(I)$.

• Observe that $g': I \to \mathbb{R}$ cannot be identically $0$ on $I$, otherwise by the Mean-Value Theorem, $g$ would be a constant function.

• Hence, there exists an $x_{0} \in I$ such that $g'(x_{0}) \neq 0$. Without loss of generality, let us assume that $g'(x_{0}) > 0$.

• As $g'$ is continuous on $I$, there exists an open subinterval $J$ of $I$ such that $x_{0} \in J$ and $g' > 0$ on $J$. Intuitively speaking, as $g'(x_{0}) > 0$, the continuity of $g'$ ensures that $g'(x) > 0$ for all points $x \in I$ that are near $x_{0}$.

• By the Mean-Value Theorem, $g|_{J}$ must be strictly increasing.

• Hence, $(g|_{J})^{-1}: g[J] \to J$ exists and is continuous on $g[J]$, which is an open interval.

• We now need to find a continuous $f: g[J] \to \mathbb{R}$ such that $$\forall x \in J: \quad g'(x) = f(g(x)).$$

• If such an $f$ exists, then it is necessary that $$\forall x \in g[J]: \quad g'({(g|_{J})^{-1}}(x)) = f(g({(g|_{J})^{-1}}(x))) = f(x).$$

• This implies that $f = g' \circ (g|_{J})^{-1}$, which is continuous on $g[J]$ because it is the composition of two continuous functions.

• A quick check shows that $f$ as defined does satisfy the requirements.

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