Central bank loss function
The intuitive way


Draw the function $L_t$, the graph below is build with $\gamma = 0.8$ and $\pi_t^\otimes = 1$



Convex in this context means that the function $L_t$ looks like a bowl, just one well defined minimum. And if you cut it with a horizontal plane, the result is an ellipse.




  The formal way


Just for convenience, define the variable $p_t = \pi_t - \pi_t^\otimes$ so the function becomes

$$
L_t(p_t, Y_t) = \gamma p_t^2 + Y_t^2 \tag{B1}
$$

First note that if $\gamma &gt; 0$  then $L_t(p_t, Y_t) \ge 0$. To prove convexity you just need to show that for any pair $(p_{t,1},Y_{t,1})$ and $(p_{t,2},Y_{t,2})$ the condition

$$
L_t(t p_{t,1} + (1 - t)p_{t,2}, t Y_1 + (1 - t)Y_{t,2}) &lt; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \tag{B2}
$$

holds for $ 0 &lt; t &lt; 1$. This is actually fairly easy to show

\begin{eqnarray}
L_t(t p_{t,1} + (1 - t)p_{t,2}, t Y_1 + (1 - t)Y_{t,2}) &amp;=&amp; \gamma[t p_{t,1} + (1 - t)p_{t,2}]^2 + (t Y_1 + (1 - t)Y_{t,2})^2 \\
&amp;\vdots &amp; \\
&amp;=&amp; t [\gamma p_{t,1}^2 + Y_{t,1}^2] + (1 - t) [\gamma p_{t,2}^2 + Y_{t,2}^2] \\
&amp;&amp;\quad + t(1 - t) [\gamma(p_{t,1} - p_{t,2})^2 + (Y_{t,1} - Y_{t,2})^2] \\
&amp;=&amp; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \\
&amp;&amp;\quad \underbrace{t(1 - t)L_t(p_{t,1} - p _{t_2}, Y_{t,1} - Y_{t,2})}_{&gt; 0} \\
&amp;&lt; &amp; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \tag{B3}
\end{eqnarray}

So $L_t$ is convex. To show that indifference curves are ellipses, just fix $L_t$ to a constant value, let's say $A &gt; 0$ then we have

$$
A = \gamma p_t^2+ Y_t^2 ~~\Rightarrow~~ \frac{p_t^2}{A/\gamma} + \frac{Y_t}{A} = 1 \tag{B4}
$$

which are just ellipses centered at $p_t = 0 = \pi_t - \pi_t^\otimes$ and $Y_t = 0$ with axis lengths $\sqrt{A/\gamma}$ and $\sqrt{A}$