Show $f(x)=x^TBx$ is not uniformly continuous if $B$ is symmetric

Question

I am in the initial fase of learning analysis. I am working on the following problem:

Consider the quadratic form $f \, : \, \mathbb{R}^k \to \mathbb{R} $ given by $f(x)=x^TBx$ for $x \in \mathbb{R^k}$ and a $k \times k$ matrix $B$.

1. Show that if there exists $x_0 \in \mathbb{R}^k$ such that $x_0^TBx_0 \neq 0$, then $f$ is not uniformly continuous.
   Hint: Consider $t \mapsto f(tx_0)$.

2. Show that if $B$ has a diagonal element $b_{ii}$ such that $b_{ii} \neq 0$, then $f$ is not uniformly continuous.

3. Show that if $B$ is symmetric and different from the zero matrix, then $f$ is not uniformly continuous.

4. Find an example of a matrix $B$ different from the zero matrix such that $f$ is uniformly continuous.

Regarding question 4, I figured out that the uniform continuity is not ruled out by 1-3 in the example where B is a 2x2 matrix with zeros in the diagonal and the other two elements equal to $a$ and $−a$. And then the function $f$ maps every element to zero, so this function is clearly uniformly continuous.

But I am having trouble with the first 3 questions. Any hints would be greatly appreciated!

Answer 1

(1) Prove by contradiction. Suppose the contrary that $f$ is uniformly continuous. For $\epsilon=1$, there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $||x-y||<\delta$. Note that $f(tx)=t^{2}f(x)$ for any $t\in\mathbb{R}$ and $x\in\mathbb{R}^{k}$. Choose $t_{1}>\frac{2||x_{0}||}{|f(x_{0})|\delta}.$ Define $t_{2}=t_{1}+\frac{\delta}{2||x_{0}||}$ (Note that $x_{0}\neq0$, so $||x_{0}||>0).$ Observe that $||t_{1}x_{0}-t_{2}x_{0}||=||x_{0}||\cdot|t_{2}-t_{1}|=\frac{\delta}{2}<\delta$, so we have $\left|f(t_{1}x_{0})-f(t_{2}x_{0})\right|<\epsilon=1.$ On the other hand, by direct calculation, \begin{eqnarray*} \left|f(t_{1}x_{0})-f(t_{2}x_{0})\right| & = & |f(x_{0})||t_{1}^{2}-t_{2}^{2}|\\ & = & |f(x_{0})|\cdot(t_{2}-t_{1})(t_{2}+t_{1})\\ & \geq & |f(x_{0})|\cdot\frac{\delta}{2||x_{0}||}\cdot2t_{1}\\ & \geq & |f(x_{0})|\cdot\frac{\delta}{2||x_{0}||}\cdot\frac{4||x_{0}||}{|f(x_{0})|\delta}\\ & = & 2, \end{eqnarray*} which is a contradiciton.

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(3) follows from the fact : If $B$ is a symmetric $k\times k$ matrix and $\langle Bx,x\rangle=0$ for all $x\in\mathbb{R}^{k}$, then $B=0$.

Proof: Let $x,y\in\mathbb{R}^{k}$. Since $B$ is symmetric, we have that $\langle By,x\rangle=\langle y,B^{t}x\rangle=\langle y,Bx\rangle=\langle Bx,y\rangle.$ Therefore \begin{eqnarray*} 0 & = & \langle B(x+y),(x+y)\rangle\\ & = & \langle Bx,x\rangle+\langle By,y\rangle+\langle Bx,y\rangle+\langle By,x\rangle\\ & = & 0+0+2\langle Bx,y\rangle. \end{eqnarray*} That is, $\langle Bx,y\rangle=0$. It follows that $B=0$.

Therefore, if $B$ is symmetric and $B\neq0$, there exists $x_{0}$ such that $\langle Bx_{0},x_{0}\rangle\neq0$.

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(4) Simply find a non-zero $k\times k$ matrix such that $B+B^{t}=0.$ Such matrix exists. For example, let $B=(b_{ij})$ with $b_{12}=-1$, $b_{21}=1$, and $b_{ij}=0$ for other entries. In this case $\langle Bx,x\rangle=\langle x,B^{t}x\rangle=\langle x,-Bx\rangle=\langle-Bx,x\rangle=-\langle Bx,x\rangle$ and it will follow that $\langle Bx,x\rangle=0$ for all $x$.