If $f, g$ are both smooth functions of $t$, why does $f \circ g$ is a smooth function of $t$?

Question

Let $f,g: \mathbb{R} \to \mathbb{R}$.

I think it is clear that $f \circ g$ has derivatives of all orders since $\dfrac{d^n}{dt^n} (f\circ g)$ only depends of $f, g$ and it's derivatives. But I'm asking for a formal proof because I can't figure it out one.

Can someone helps with a formal proof?

Thanks.

Answer 1

We can prove the following statement by induction on $n$: for all $n\in\mathbb{N}$, if $f,g:\mathbb{R}\to\mathbb{R}$ are $n$ times differentiable, then $f\circ g$ is $n$ times differentiable. The base case $n=0$ is trivial.

Now suppose $n>0$ and the statement is known for $n-1$ and we wish to prove it for $n$. Let $f,g:\mathbb{R}\to\mathbb{R}$ be $n$ times differentiable. By the chain rule, $f\circ g$ is differentiable with $(f\circ g)'=g'\cdot (f'\circ g)$. Now $f'$ and $g$ are both $n-1$ times differentiable, so by the induction hypothesis, $f'\circ g$ is $n-1$ times differentiable. Since $g'$ is also $n-1$ times differentiable, $(f\circ g)'$ is $n-1$ times differentiable since it is a product of two $n-1$ times differentiable functions. Thus $f\circ g$ is $n$ times differentiable.

(Here I assume you already know that a product of two functions which are $n$ times differentiable is $n$ times differentiable. If you don't know that, it can be proved by induction on $n$ in the same way, using the product rule instead of the chain rule.)

Answer 2

As $f$ is cont., we see that $f(x+t)\rightarrow f(x)$ as $t\rightarrow 0$. So, we have

$$\lim_{t\rightarrow 0}\frac{(g\circ f)(x+t)-(g\circ f)(x)}{t} =\lim_{f(x+t)\rightarrow f(x)}\frac{g(f(x+t))-g(f(x))}{t}$$

But, we don’t know what is this. Arrange this as,

$$\lim_{f(x+t)\rightarrow f(x)}\frac{g(f(x+t))-g(f(x))}{t} =\lim_{f(x+t)\rightarrow f(x)}\frac{g(f(x+t))-g(f(x))}{f(x+t)-f(x)}\cdot\frac{f(x+t)-f(x)}{t}$$

Limit distributes in product (here) and so, we have $$\lim_{f(x+t)\rightarrow f(x)}\frac{g(f(x+t))-g(f(x))}{f(x+t)-f(x)}\lim_{t\rightarrow 0}\frac{f(x+t)-f(x)}{t}=g’(f(x))f’(x)$$

So, the limit exists and so the composition $g\circ f$ is differentiable.

Suppose $g,f$ are smooth. It is clear that $g\circ f$ is differentiable.

$g’\circ f$ is differentiable being composition of differentiable functions.

As product of differentiable functions is a differentiable function, $(g’\circ f)\cdot f’$ is differentiable, that is, $(g\circ f)’$ is differentiable.

Similarly, we can prove every derivative of $g\circ f$ is differentiable, that is $g\circ f$ is smooth.