I want to show that the following function is convex without using the Hessian-based definition of convex.
$$ f(x) = \| x - z \|_2 $$
I believe I am supposed to use this definition:
$$ f(\alpha x + (1-\alpha)y) \leq \alpha f(x) + (1-\alpha)f(y) $$
Doing so for the function $f(x) = \| x \|_2$ is trivial it seems: $$ \begin{align*} \| \alpha x + (1-\alpha)y \|_2 &\leq \| \alpha x \|_2 + \| (1-\alpha) y \|_2 \text{ by triangle inequality} \\ &= \alpha \| x \|_2 + (1-\alpha) \| y \|_2 \text { by homogeneity of } L^2 \text{ norm} \end{align*} $$ This satisfies the definition of convex.
When I try to apply this logic to the same function, this is as far as I can get: $$ \begin{align*} \| \alpha x + (1-\alpha) y - z \|_2 &\leq \alpha \| x \|_2 + \|(1-\alpha)y - z \|_2 \end{align*} $$ using the same triangle inequality as above. How do I handle the $-z$ term?
Depending on how you're denoting the $L^2$-norm, you might need to redo the proof for $\|x\|_2^2$ (this looks like the square of the $L^2$-norm to me), but after you have that, the result for the shifted function follows from $$\alpha x + (1-\alpha)y - z = \alpha(x-z) + (1-\alpha)(y-z)$$