# Proving convexity of $f$ , $f(z)<f(x)$ [closed]

Prove that if $$f$$ is convex, $$f(z) < f(x)$$, $$g ∈ ∂f(x)$$, then for small $$t > 0$$: $$‖x − tg − z\|^2 ≤ ‖x − z‖^2.$$

It's equivalent to show $$\Vert x − tg − z\Vert_2^2 ≤ \Vert x − z\Vert_2^2$$,

That is to show $$x^Tx+z^Tz+tg^Ttg-2x^Tz-2x^Ttg+2z^Ttg\le x^Tx+z^Tz-2x^Tz$$

That is to show $$tg^T(tg+2(z-x))\le 0$$.

since $$t>0$$, that is to show $$tg^Tg+2g^T(z-x)\le 0$$ (1).

By the definition of subgradient, $$\forall y, f(y)\ge f(x)+g^T(y-x)$$.

So $$f(z)\ge f(x)+g^T(z-x) \Rightarrow 0>f(z)-f(x)\ge g^T(z-x)$$

Go back to (1) , since $$2g^T(z-x)<0$$, one can always find a small $$t$$ such that (1) is satisfied.