Show there exists a constant $C>0$ such that $\forall x\in\mathbb{R}^{n},x^{t}Ax\leq C\cdot\left\Vert x\right\Vert ^{2}$

Question

Question

Let $A\in M_{n\times n}\left(\mathbb{R}\right)$ be a square symmetric matrix. Show that there exists constant $C>0$ such that for any $x\in\mathbb{R}^{n},x^{t}Ax\leq C\cdot\left\Vert x\right\Vert ^{2}$.

In addition, find the best possible constant $C$ for which the inequality holds.

Guidance:

Write down the left hand side explicitly, and use the inequality $\left|ab\right|\leq\frac{\left(a^{2}+b^{2}\right)}{2}$.

How can I use the guidance in order to solve this?

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Edit

First part (existence):

I think I have managed to solve the first part as follows (with the help of Robert Israel's comment):

\begin{aligned}x^{t}Ax=\sum_{i\in\left[n\right]}\sum_{j\in\left[n\right]}x_{i}a_{ij}x_{j}=\sum_{i\in\left[n\right]}\sum_{j\in\left[n\right]}a_{ij}x_{i}x_{j} & =\sum_{j\neq i\in\left[n\right]}a_{ij}x_{i}x_{j}+\sum_{\begin{array}{c} i=j\in\left[n\right]\\ x_{i}\neq0 \end{array}}a_{ii}x_{i}^{2}\\ & =\sum_{\begin{array}{c} i=j\in\left[n\right]\\ x_{i}\neq0 \end{array}}x_{i}^{2}\left(\left(\sum_{\begin{array}{c} j\neq k\in\left[n\right]\\ k,j\neq i \end{array}}\frac{a_{jk}x_{j}x_{k}}{x_{i}^{2}}\right)+a_{ii}\right)\\ & \underbrace{\leq}_{\text{Guidance}}\sum_{\begin{array}{c} i=j\in\left[n\right]\\ x_{i}\neq0 \end{array}}x_{i}^{2}\underbrace{\left(\left(\sum_{\begin{array}{c} j\neq k\in\left[n\right]\\ k,j\neq i \end{array}}\frac{\left|a_{jk}\right|\left(x_{j}^{2}+x_{k}^{2}\right)}{2x_{i}^{2}}\right)+\left|a_{ii}\right|\right)}_{C>0}\\ & =\left\Vert x\right\Vert ^{2}C \end{aligned}

Second part (finding the best):

Assuming the C above exists, this is how I have tried using orthogonal diagonalization in order to solve this:

\begin{aligned}x^{t}Ax & =x^{t}QDQ^{t}x\\ & =\left(Q^{t}x\right)^{t}DQ^{t}x\\ & =\sum_{i=1}^{n}\lambda_{i}\left(Q^{t}x\right)^{2}\\ & \leq\sum_{i=1}^{n}\lambda_{max}\left(Q^{t}x\right)^{2}\\ & =\lambda_{max}\left\Vert Q^{t}x\right\Vert ^{2}\\ & =\lambda_{max}\left\Vert x\right\Vert ^{2} \end{aligned}

when $A=QDQ^{T}$.

However, I couldn't find a way to show this is the best C.

How can I prove it?

Answer 1

The left side is a sum of terms $a_{ij} x_i x_j$. By your inequality, that $\le |a_{ij}| (x_i^2 + x_j^2)/2$. Add those up, and you get a sum of constants times $x_i^2$. Don't worry about what constants, if all you need is the existence of $C$.

For the second part you'll want to use the orthogonal diagonalization.