Analytic gradient $\nabla f$ implies analytic $f$

Question

I'm wondering how to prove that a real function $f : \mathbb{R}^d \to \mathbb{R}$ with analytic gradient $\nabla f$ (equivalently, analytic Fréchet derivative $Df$) must also be analytic. We can simply take the antiderivative of the Taylor expansion of $f'$ if $d=1$, as shown here, but what about $d > 1$?

Answer 1

Analyticity of $f:\mathbb{R}^n\to\mathbb{R}$ at $0\in\mathbb{R}^n$ is equivalent to saying that $$ |{\partial^\alpha f}(0)|\leq M\frac{\alpha !}{r^{|\alpha|}} , $$ for some constants $M$ and $r>0$, where $\alpha=\alpha_1!\cdots\alpha_n!$ and $|\alpha|=\alpha_1+\ldots+\alpha_n$ for a multi-index $\alpha\in\{0,1,\ldots\}^n$. Since $\partial_kf$ is analytic, we have $$ |{\partial^\alpha \partial_kf}(0)|\leq M_k\frac{\alpha !}{r_k^{|\alpha|}} , $$ for some constants $M_k$ and $r_k>0$, $k=1,\ldots,n$. From this, it is almost trivial to conclude that $$ |{\partial^\alpha f}(0)|\leq M\frac{\alpha !}{r^{|\alpha|}} , $$ where $$ M=\max\{|f(0)|,r_1M_1,\ldots,r_nM_n\}, \qquad\textrm{and}\qquad r=\min\{r_1,\ldots,r_n\}. $$ This implies that $f$ is analytic at $0$. Now since $0$ was an arbitrary point we can conclude that $f$ is analytic everywhere.

In fact, this argument shows that a sufficient condition for analyticity of $f$ is that $\partial_1f$ is analytic and the gradient $\nabla f$ exists.

Answer 2

I am sure there are simpler arguments but one way would be to observe that $\Delta f$ is analytic. Then $f$ must be analytic by elliptic regularity.

Answer 3

Take $d=2$ for example. To show $f$ is analytic near $(0,0),$ first note $D_1f(x,y)$ has a power series expansion

$$ \sum c_{mn}x^my^n$$

that converges absolutely and uniformly in a nighborhood of $(0,0).$ It follows that for $(x,y)$ close to $(0,0),$

$$f(x,y) = f(0,y) + \int_0^x D_1f(t,y)\,dt $$ $$=\int_0^x \left (\sum c_{mn}t^my^n\right )\,dt = f(0,y) + \sum c_{mn}\frac{t^{m+1}}{m+1}y^n.$$

But as you mentioned, the result is true for $d=1.$ Thus $f(0,y)$ equals a power series in $y$ for $y$ close to $0.$ It follows that $f(x,y)$ equals a power series in a neighborhood of $(0,0).$

I took $(0,0)$ as the base point for convenience; any base point could have been taken. Also, the argument can be extended to higher dimensions using induction.