can anybody help me with this question? Let $A\in R^{nxn}$ a positive semidefinite matrix and $\alpha \geq 0 $ I want to prove the following:
$ M_{\alpha} = { x \in R^n | x^{T}Ax \leq \alpha}$ is convex? to prove that I am adviced to use the following tip after proving it for any $ \lambda,\mu \in R $ the following holds $\lambda^2x^{T}Ax+2\lambda\mu x^T A y+ \mu^2 y^{T}Ay \geq 0$
Please excuse me for the bad writing
If you take $x_1,x_2\in M_\alpha$, you have to show that for $t\in[0,1]$, such that , $tx_1+(1-t) x_2 \in M_\alpha$.
Then, you can compute : $$(tx_1+(1-t)x_2)^TA(t x_1+(1-t) x_2)=t^2x_1^TAx_1+(1-t)^2x_2^TAx_2+t(1-t)(x_1^TAx_2+x_2^TAx_1)\\ = t^2x_1^TAx_1+(1-t)^2x_2^TAx_2+t(1-t)\left (x_1^T\left(\frac{x_2x_2^T}{||x_2||^2}\right)Ax_2+x_2^T\left(\frac{x_1x_1^T}{||x_1||^2}\right)Ax_1\right) \\ \leq t^2x_1^TAx_1+(1-t)^2x_2^TAx_2+t(1-t)\alpha\left(\frac{x_1^Tx_2}{||x_2||^2}+\frac{x_2^Tx_1}{||x_1||^2}\right) \\ \leq \left(t^2+(1-t)^2+t(1-t)<x_1,x_2>^2\left(\frac{||x_1||^2+||x_2||^2}{(||x_1||\cdot||x_2||)^2} \right)\right)\alpha\text{ (using Cauchy-Schwarz)}\\ \leq \left(t^2+(1-t)^2+t(1-t)(||x_1||^2+||x_2||^2)\right)\alpha\\ \leq \left(1-\frac{||x_1||^2+||x_2||^2}{2} \right)\alpha$$
Well, I can't get to the final result, but with this, I can get that $M_\alpha$ is convex on the unitary ball on $\mathbb{R}^n$