Show that $g \circ f$ is n times differentiable

Question

Let $D,E \subseteq \mathbb{R}$ and $f: D \to E, g: E \to \mathbb{R}$ be two $n$ times differentiable functions. Then $g \circ f: D \to \mathbb{R}$ is $n$ times differentiable.

I was thinking about using induction over $n$. I know the base case with $n=1$ is true and assume it is true for $n$. But I struggle to show the induction step $(g \circ f)^{(n+1)} := \Big( (g \circ f)^{(n)} \Big)' = \ldots$

If possible I want to avoid using the concrete formula for the n-th chain rule because it is quite complicated.

Answer 1

You can prove by induction on the times $f$ and $g$ are differentiable:

1. If $f:D\to E$ and $g:E\to \mathbb{R}$ are differentiable (once), then we already know that (it's a fundamental theorem) $$ (g \circ f)' = g'\circ f \cdot f'. $$
2. Now suppose that $f$ and $g$ are $n$-times differentiable, since they are also $n-1$-times differentiable we can use the induction step and say that $g\circ f$ is $n-1$-times differentiable, and holds the formula $$ (g\circ f)^{n-1} = \sum_{\beta\leq n-1} g^{\beta}\circ f \cdot f^{[n,m]_{\beta}}, $$ where $[n,m]_\beta$ stands for a $n$-th polynomial of the derivatives up to order $m$-th (just a made up notation). Now all the term into the expression above are either polynomials, derivative of order $\beta, m \leq n-1$ or composition of smooth functions, so that we can use the same theorem that we recalled above, the fact that polynomials are smooth and the fact that products of smooth functions are differentiable to deduce that $(g\circ f)^{n-1}$ is differentiable, i.e $g\circ f$ is $n$-times differentiable.

Answer 2

Hint For $h = g \circ f$, you have $h^\prime(x) = g^\prime(f(x)) \circ f^\prime(x)$.

Now consider the two maps: $\phi_1: x \mapsto (g^\prime(f(x)), f^\prime(x))$ and $\phi_2: (u, v) \mapsto u \circ v$ - function composition of linear maps.

You have $h^\prime(x) = (\phi_2 \circ \phi_1)(x)$. $\phi_2$ is bilinear, hence $\mathcal C^\infty$.

Based on that, you can proceed by induction. If $f,g \in \mathcal C^n$, $f^\prime, g^\prime \in \mathcal C^{n-1}$ and by induction $\phi_1 \in \mathcal C^{n-1}$. Therefore $h^\prime \in \mathcal C^{n-1}$ and $h \in \mathcal C^n$.

Answer 3

Let the order of an expression involving derivatives be the highest derivative appearing in that expression.

Then the order of $$(fog)'(x) = f'(g(x)).g'(x)$$ is $1.$

We prove by induction on the order of $(fog)^{(n)}(x)$

For $n=1$ the right side of $(fog)'(x) = f'(g(x)).g'(x)$ has order $1$ therefore the left side is well defined.

Note that $$(fog)''(x) = f''(g(x)).(g'(x))^2 + g''(x).f'(g(x))$$ has order $2$ and it is well defined because all the functions involved are twice differentiable.

Assume that $(fog)^{(k-1)}(x)$ is an expression of order $k-1$ which is well defined.

Differentiation of such an expression results in an expression of order k, which is well defined because each term in that expression is differentiable.