Show that the given set is convex for $a,b \in \mathbb{R}$ 
Show for arbitrary $a,b \in \mathbb{R}$ that the given set is
  convex$$E(a,b):=\left\{\begin{pmatrix} x_1\\  x_2 \end{pmatrix} \in
\mathbb{R}^2 \, \Bigg\vert \, \frac{x^2_1}{a^2} + \frac{x^2_2}{b^2}\leq
1\right\}$$

I think because the set is written in terms of vectors, I need to use the vector definition which is, if I understood correctly (sadly there was no definition in my books):


*

*A set $S$ is convex if for any $u,v \in S$, the point $tu + (1-t)v \in S$ for all $t \in [0,1]$
So then let for the example above $x_1, z \in E$ and $t \in [0,1]$. We have
$$\frac{(tx_1+(1-t)z)^2}{a^2}+\frac{x_2^2}{b^2} = \\ = \frac{t^2x_1^2}{a^2}-\frac{2t^2x_1z}{a^2}+\frac{t^2z^2}{a^2}+\frac{2tx_1z}{a^2}-\frac{2tz^2}{a^2}+\frac{z^2}{a^2} +\frac{x_2^2}{b^2}$$
Now I'm not sure how to continue because is really every fraction lower/equal than $1$? The second and fifth fraction surely are because they are negative, the last fraction must be lower/equal than $1$ as well because it's from the original set...
I somehow wanted continue by writing an inequality now and forming for $t$ but no idea how to complete this? Maybe there is another way of showing this easier?
 A: Hint: $x \to x^{2}$ is a convex function, so $(tx_1+(1-t)y_1)^{2} \leq tx_1^{2}+(1-t)y_1^{2}$. Write a similar inequality for second coordinates and 'combine'. [I am writing $u$ as $(x_1,x_2)$ and $v$ as $(y_1,y_2)$]
A: I don't really get what are $x_1$ and $z$ in your demonstration. You say they're elements of E, which is a subset of $\mathbb R^2$, but then you treat them as elements en $\mathbb R$.
So let's get back to the definition you used:

A set $S$ is convex if for any $u,v\in S$, the point $tu+(1−t)v \in S$ for all $t \in [0,1]$

From here, suppose $u$ and $v$ in $E$. Which means:
$$\frac{u_1^2}{a^2}+\frac{u_2^2}{b^2} \le 1 $$
$$\frac{v_1^2}{a^2}+\frac{v_2^2}{b^2} \le 1 $$
What you have to prove is that $tu+(1−t)v \in E$. So you need:
$$\frac{(tu_1+(1-t)v_1)^2}{a^2}+\frac{(tu_2+(1-t)v_2)^2}{b^2} \le 1 $$
You can now use Kavi Rama Murthy's answer to finish the problem
A: we have $x^TAx \le 1$  and $y^TAy \le 1$ where $A$ is positive definite and also diagonal.
\begin{align}
&(t x + (1-t) y)^TA(t x + (1-t)y)\\
&= t^2 x^TAx+2t (1-\lambda) x^TAy + (1-t)^2y^TAy \\
&\le t^2 x^TAx+2t (1-\lambda) \sqrt{(x^TAx)(y^TAy)} + (1-t)^2y^TAy \\
&=(t \sqrt{x^TAx}+(1-t)\sqrt{y^TAy})^2\\
&\le 1
\end{align}
