Upper/Lower bounds of real-analytic functions with infinite Taylor series

Question

For example, in 1-D, given some positive increasing polynomial $p(x) = a_1x+\ldots+a_nx^n$, $p(0) = 0$, there exists constants $b_1,b_2$ such that for $x<\delta$, for some $\delta > 0$, we have that $$ b_1 x^{n_1} \leq p(x) \leq b_2 x^{n_1}, $$ where $n_1$ is the smallest exponent on $x$. Additionally, the same property holds for real-analytic functions.

I was wondering if there's an 2-D analogue of this property in regards to real-analytic functions that do not have a finite Taylor series expansion. Additionally, lets assume that these functions are strictly increasing along lines intersecting the origin.

If my function is a polynomial it is pretty straightforward. For example, consider $f(x,y) = x^2 + (x^2+y^2)^2$. It follows that $$ (x^2+y^2)^2 \leq f(x,y) \leq 2 (x^2+y^2). $$ for $(x,y)\in B_{\delta}$ for some $\delta > 0$. The lower bound is determined by the largest exponent and the upper bound is determined by the smallest exponent.

Now, if $f(x,y)$ is instead defined in terms of some infinite Taylor series can one still find such inequalities? Intuitively, it feels like the upper bound could still be found, but a lower bound seems far more difficult. Any advice would be appreciate.

Answer 1

You can use the Weierstrass preparation theorem and recursively apply the 1-D result. The theorem asserts that an analytic function $f(x,\mathbf{y})$, for which $f(0,0)=0$, can be written as a product $f(x,\mathbf{y}) = W(x,\mathbf{y}) g(x,\mathbf{y})$, where $W(0,0) \ne 0$ and $$ g(x,\mathbf{y}) = x^k + g_{k-1}(\mathbf{y}) x^{k-1} + \cdots + g_0(\mathbf{y}) $$ is a Weierstrass polynomial of some degree $k$, with analytic coefficients such that $g_i(\mathbf{y}=0) = 0$.

By continuity, there is some $\delta > 0$ such that $0 < C_1 \le W(x,\mathbf{y}) \le C_2$ on $(x,\mathbf{y}) \in B_\delta$, so $$ C_1 g(x,\mathbf{y}) \le f(x,\mathbf{y}) \le C_2 g(x,\mathbf{y}) $$ reduces the bounds to those on $g(x,\mathbf{y})$. If we have bounds on the coefficients $b_i(\mathbf{y}) \le g_i(\mathbf{y}) \le c_i(\mathbf{y})$ for $|\mathbf{y}| \le \delta$, then $$ x^k + b_{k-1}(\mathbf{y}) x^{k-1} + \cdots + b_0(\mathbf{y}) \le g(x,\mathbf{y}) \le x^k + c_{k-1}(\mathbf{y}) x^{k-1} + \cdots + c_0(\mathbf{y}) . $$ To get the coefficient bounds, you can now apply the above steps recursively, possibly shrinking $\delta$ as necessary. Once you get to the 1-D case, you can terminate by using the bounds given by monomials. The result will be of the form $$ b(x,\mathbf{y}) \le f(x,\mathbf{y}) \le c(x,\mathbf{y}) , $$ where $b$ and $c$ are polynomials. If your original function only had positive Taylor coefficients, I suspect that all the intermediate upper and lower bounds should also have positive coefficients, but I'm not sure. Perhaps some other trick could be used if a negative lower bound appears for one of the coefficients at an intermediate step.

The lower bound $b$ is in general not going to be a monomial, but if it does have only positive coefficients, you could drop from it as many terms as you would like. Applying this idea to your example \begin{align*} f(x,y) &= x^4 + 2 (y^2 + 1) x^2 + y^4 \\ &= W(x,y) \left(x^2 + \frac{1}{2} + 2y^2 - \sqrt{y^2 + \frac{1}{4}}\right) \\ &= W(x,y) (x^2 + y^4 - 2 y^6 + \cdots) \end{align*} gives the bounds $$ B (x^2 + y^4) \le B_2 (x^2 + B_0 y^4) \le f(x,y) \le C_2 (x^2 + C_0 y^4) \le C (x^2 + y^4) . $$ The outer bounds follow from the inner ones by setting for instance $B = \min(B_2, B_2 B_0)$ and $C = \max(C_2, C_2 C_0)$. Switching around the roles of $x$ and $y$, $f(x,y) = y^4 + 2 x^2 y^2 + (x^4 + x^2)$ is already in Weierstrass form and the resulting bounds are $$ B (y^2 + x^2)^2 \le f(x,y) \le C (y^4 + x^2 y^2 + x^2) . $$ The middle coefficient in the upper bound could also be any other positive constant.

Answer 2

It seems you are asking for a non-negative lower bound that has an isolated zero at the origin: recall that even if $f(0)=0$ and $f$ is strictly increasing along every radial direction, it may vanish e.g. on a parabola, like $f(x,y):=(y-x^2)^2$, so that one can’t put anything like $b(x^{2m}+y^{2m})$ below it in a nbd of the origin, as you were asking in a comment.