derivative of the functions in Hadamard's lemma

Question

I think Corollary 1.10 of this PDF page 12 is Hadamard's lemma here $\partial_jf$ is the partial derivative $\frac{\partial f}{\partial x_j}$ and $\alpha$ is a multi-index. $\partial^\alpha$ is defined by $$\partial^\alpha u(x):=\frac{\partial^{|\alpha|} u}{\partial x_1^{\alpha_1} \ldots \partial x_n^{\alpha_n}}(x)$$ The proof of Corollary 1.10 (on the next page) defines the functions $$f_j(x)=\int_0^1\left(\partial_j f\right)\left(x_0+t\left(x-x_0\right)\right) \mathrm{d} t$$ It says The assertions about $f_j$ all follow by inspection.
I don't know how to deduce one of the assertions: $$\partial^\alpha f_j\left(x_0\right)=\partial^\alpha \partial_j f\left(x_0\right) /(1+|\alpha|)$$ where does the factor $1/(1+|α|)$ come from?

Answer 1

Plugging in $f_j$ we need to prove: [At this point we cannot replace $x$ with $x_0$ in the integral, because the integral has $\partial^\alpha$ in front of it, we need to evaluate the derivative of a function of $x$ at $x=x_0$, if we had replaced $x$ with $x_0$ in the integral, then it would be the derivative of a constant.] $$\partial^\alpha \int_0^1(\partial_j f)\left(x_0+t\left(x-x_0\right)\right) \mathrm{d} t\Bigg|_{x=x_0}=\partial^\alpha \partial_j f\left(x_0\right) /(1+|\alpha|)$$ Interchanging $\partial^\alpha \int_0^1$ $$\int_0^1\partial^\alpha\Bigl((\partial_j f)\left(x_0+t\left(x-x_0\right)\right)\Bigr)\mathrm{d} t\Bigg|_{x=x_0}=\partial^\alpha \partial_j f\left(x_0\right) /(1+|\alpha|)$$ Applying chain rule $$\int_0^1t^{|\alpha|}(\partial^\alpha\partial_j f)\left(x_0+t\left(x-x_0\right)\right)\mathrm{d} t\Bigg|_{x=x_0}=\partial^\alpha \partial_j f\left(x_0\right) /(1+|\alpha|)$$ Replacing $x$ with $x_0$ in the integral $$\int_0^1t^{|\alpha|}(\partial^\alpha\partial_j f)\left(x_0+t\left(x_0-x_0\right)\right)\mathrm{d} t=\partial^\alpha \partial_j f\left(x_0\right) /(1+|\alpha|)$$ Canceling $x_0-x_0$ $$\int_0^1t^{|\alpha|}(\partial^\alpha\partial_j f)\left(x_0\right)\mathrm{d} t=\partial^\alpha \partial_j f\left(x_0\right) /(1+|\alpha|)$$ Note that $(\partial^\alpha\partial_j f)\left(x_0\right)$ in the integral is constant wrt $t$, and $\int_0^1t^{|\alpha|}\mathrm dt=\frac1{1+|\alpha|}$, we are done.