I'm solving the following problem.
Problem. Let $A(x)$ be a $n\times n$ real symmetric matrix on $\mathbb{R}^n$ with $$ |((A(x)\xi) \cdot \xi | \leq C|\xi|^2.$$ Then $A$ is bounded.
Here is my proof. Fix $x\in \mathbb{R}^n$. Since $A(x)$ is real and symmetric, by spectral theorem, there exists an orthonormal basis for $A(x)$ so that we can diagonalize it. In particular, we can represent $A(x) = P^T D(x) P$, where $D(x)$ is a diagonal matrix whose entries are eigenvalue of $A$.
Now from the condition, if $\mathbf{v}$ is an unit eigenvector with eigenvalue $\lambda (x)$, then we get $|\lambda (x)|\leq C$. This implies every eigenvalue of $A$ is bounded.
Now note that $$ \Vert A \Vert_2 =\sqrt{\lambda_\max (A^TA)}=\sqrt{\lambda_\max (D^2)}$$ since $A^T A =P^T D^2 P$ and similar matrices have same eigenvalues.
Also note that $$ \Vert{A(x)}\Vert_F \leq \sqrt{n} \Vert A(x) \Vert_2\leq C^\prime,$$ where $$ \Vert A \Vert_F = \left(\sum_{i,j=1} a_{ij}^2\right)^{\frac{1}{2}}.$$
This shows $A$ is bounded.
But the proof requires several concepts of matrix norm. How can I prove it more easily?
Let $e_1,\dots,e_n$ denote the standard basis. If we set $\xi = e_i$, the inequality yields $$ |A_{ii}(x)| \leq C $$ If we set $\xi = e_i + e_j$ for $i \neq j$, the inequality yields $$ |A_{ii}(x) + 2A_{ij}(x) + A_{jj}(x)| \leq 2C $$ Using these two inequalities, we have (for $i \neq j$) $$ |A_{ij}| = \left|A_{ij} + \frac 12 \left(A_{ii} + A_{jj}\right) - \frac 12 \left(A_{ii} + A_{jj}\right) \right|\leq\\ \frac 12 \left|A_{ii} + 2A_{ij} + A_{jj}\right| + \frac 12 \left|A_{ii} + A_{jj}\right| \leq \\ \frac 12 \left|A_{ii} + 2A_{ij} + A_{jj}\right| + \frac 12 (|A_{ii}| + |A_{jj}|) \leq\\ C + C = 2C $$ We may conclude that $A_{ij} \leq 2C$ for all $i$ and $j$, from which it follows that $$ \|A(x)\|_F \leq 2nC $$ So that $A(x)$ is bounded, as desired.