Given $h \in \mathbb R^n $ is it true that $\lim_{h \to 0 } \frac{|h^T S h|}{\|h\|}=0$

Question

Given a symmetric matrix $S $ and a vector $h =(h_1, \cdots , h_n ) \in \mathbb R^n $, is it true that

$$\lim_{h \to 0 } \frac{|h^T S h|}{\|h\|}=0$$

Expanding the numerator one may observe that it would be sufficient to check that

$$\lim _ {h \to 0 } \frac {|h_1^{i_1} h_2^{i_2 } ...h_n^{i_n }|} {h_1^2 + h_2^2 + \cdots + h_n^2 } $$

where $i_1 + i_2 + \cdots + i_n = n$

Is this true and what inequality could one use?

Thanks in advance!

Answer 1

Yes, this is true. It suffices to use the inequality $$ \lambda_{\min} h^Th \leq h^TSh \leq \lambda_{\max} h^Th $$ where $\lambda_{\min},\lambda_{\max}$ are the smallest and largest eigenvalues of $S$. This inequality holds by the Rayleigh-Ritz theorem.

Answer 2

Let $M:=\max_{ij}|S_{ij}|$ so $|h^TSh|\le M\sum_{ij}|h_ih_j|=M\Vert h\Vert^2$, so $\frac{|h^TSh|}{\Vert h\Vert}\le M\Vert h\Vert$. This $\to0$ as $h\to0$ by the squeeze theorem.