I have seen some posts proving the linear least squares convexity using the second derivative (see this) but I was trying to demonstrate its convexity with convex properties.
Below is my deduction. Having:
$$ y(x,w) = w_0 + w_1x + w_2x^2+...+w_Mx^M $$ $$ E(w) = \frac{1}{2}\sum_{i=1}^{N} [y(x_i,w) - t_i]^2 $$
Composition with an affine mapping is convex (if $f(z)$ is convex $f(Az+b)$ is convex). As $y(x,w) - t$ is linear on $w$ and $f(z)=z^2$ is convex. Then $[y(x,w) - t]^2$ is convex.
If $f(z)$ and $g(z)$ are convex then $f(z) + g(z)$ is convex. So $\sum_{i=1}^{N} [y(x_{i},w) - t_i]^2$ is convex.
If $f(z)$ is convex, $cf(z)$ is convex for $c>0$. Then $\frac{1}{2}\sum_{i=1}^{N} [y(x_{i},w) - t_i]^2$ is convex.
Is this correct?