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I have a question about linear approximation in the multivariate case.\

Let $f:B^d_r\to \mathbb{R}$ be a real-valued $C^2$-function defined on the $d$-dimensional ball of radius $r$ centered at the origin. Assume further that $\frac{\partial ^2f(\mathbf{x})}{\partial x_i \partial x_j}<C$ for all $i,j\in\{1,2,3,...,d\}$ where $C$ is a constant.

I am interested in the max-error $\sup_{\mathbf{x}\in B^d_r}|f(\mathbf{x})-P_1(\mathbf{x})|$ where $P_1(x)$ is a first order multivariate polynomial.

I know from Taylors theorem we have

\begin{align} f(\mathbf{x})=f(\mathbf{0})+\sum_{i=1}^d\frac{\partial f}{\partial x_i}(\mathbf{0})x_i + \sum_{|\mathbf{\alpha}|=2}R_{\mathbf{\alpha}}(\mathbf{x})\mathbf{x}^{\alpha} \end{align}

where $\alpha$ is a multi-index and $|\mathbf{\alpha}|=\alpha_1+\alpha_2+...+\alpha_d$ and $\mathbf{x}^{\mathbf{\alpha}}=x_1^{\alpha_1}x_2^{\alpha_2}...x_d^{\alpha_d}$.

We can also bound $R_{\mathbf{\alpha}}(\mathbf{x})\leq \frac{1}{\mathbf{\alpha}!}C$.

Thus if we choose $P_1(\mathbf{x})=f(\mathbf{0})+\nabla f(\mathbf{0})\cdot \mathbf{x}$ then \begin{align} |f(\mathbf{x})-P_1(\mathbf{x})|\leq \sum_{|\alpha|=2}\frac{1}{\alpha !}C\mathbf{x}^{\alpha} \end{align} Im not sure how to proceed from here. Of course I can say that $\mathbf{x}^{\alpha}\leq r^2$ for each term since $|\alpha|=2$ and similarly $\frac{1}{\alpha!}\leq 1$ so I would get:

$|f(\mathbf{x})-P_1(\mathbf{x})|\leq \frac{(d+1)d}{2}Cr^2$ as there are exactly $\frac{(d+1)d}{2}$ multi-indices $\alpha$ with $|\alpha|=2$.

My questions:

  1. Can this bound be improved?
  2. Especially, is it necessary to get the dimension $d$ into the estimate?

Kind regards

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  • $\begingroup$ For $r=1$ and $x=d^{-1/2}(1,1,\ldots,1)$ you have $\sum_{j,k=1}^d |x_jx_k|= d$. There does not seem to be reasonnable bound independent of the dimension. $\endgroup$ Commented Jun 2, 2023 at 11:23
  • $\begingroup$ @JochenWengenroth Thanks, but maybe it is possible to get a linear dependence on $d$ instead of a quadratic one that I obtained above? $\endgroup$
    – Jjj
    Commented Jun 2, 2023 at 11:29
  • $\begingroup$ Did you try to maximize $\sum x_jx_k$ subject to $\sum x_j^2=1$? $\endgroup$ Commented Jun 2, 2023 at 15:07
  • $\begingroup$ I tried it now but cant manage to solve it using lagrange multipliers $\endgroup$
    – Jjj
    Commented Jun 3, 2023 at 10:23
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    $\begingroup$ Thanks! Well, then I guess (if i have done it correctly) that the maximum value is the dimension d? Doest it seem correct? $\endgroup$
    – Jjj
    Commented Jun 3, 2023 at 11:05

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