# Show that $f(x)=\frac{1}{2}\langle Ax,x\rangle - \langle b,x\rangle$ is strictly convex $f(tu + (1-t)v) < tf(u) + (1-t)f(v),\forall t\in(0,1)$

Show that $$f(x)$$ is strictly convex, i.e., if $$u,v \in R^n$$, then $$\forall t\in(0,1)$$ this is true: $$f(tu + (1-t)v) < tf(u) + (1-t)f(v)$$

Following some reading in the previous related posts: Proof for strongly convex function is strictly convex

How can I use the proof (or not use it) and show the above inequality holds where $$f$$ is $$f(x)=\frac{1}{2}\langle Ax,x\rangle - \langle b,x\rangle$$

• This is obviously not true for any function. What are your assumptions about $f$? – Adam Latosiński May 4 at 15:23
• You mean if $f$ satisfies the defintion in the linked question? \begin{align*} f(y)\geq f(x) + \langle \nabla f(x),y-x\rangle + \frac{m}{2}\|y-x\|_2^2, \end{align*} – GNUSupporter 8964民主女神 地下教會 May 4 at 15:26
• @AdamLatosiński I'm sorry, I've edited the question. – Ilan Aizelman WS May 4 at 15:32
• @GNUSupporter8964民主女神地下教會 Sorry for the misunderstaing, I've edited the question with the appropriate $f$, you can also consider it $\phi$ – Ilan Aizelman WS May 4 at 15:33
• What is the domain of $x$ and $y$? $f$ is the solution of the integral solution? – GNUSupporter 8964民主女神 地下教會 May 4 at 15:37

## 1 Answer

$$f$$ is called strictly convex function if $$\forall x_1 \neq x_2\in X ,\forall t \in (0,1): f(tx_1+(1-t)x_2)

This question was already answered at https://math.stackexchange.com/q/3198240 by Theo Bendit you only need to do the following modifications,

\begin{align*} &\lambda\|\sqrt{A} x\|^2 + (1 - \lambda)\|\sqrt{A} y\|^2 - \|\sqrt{A}(\lambda x + (1 - \lambda)y)\|^2 \\ \ge \; &\lambda\|\sqrt{A} x\|^2 + (1 - \lambda)\|\sqrt{A} y\|^2 - (\lambda\|\sqrt{A} x\| + (1 - \lambda)\|\sqrt{A} y\|)^2 \\ = \; &\lambda(1 - \lambda)(\|\sqrt{A} x\| - \|\sqrt{A} y\|)^2 > 0. \end{align*} Since this time $$1<\lambda<0$$ so it can't be zero, and since $$x \neq y$$ we get $$(\|\sqrt{A} x\| - \|\sqrt{A} y\|) \neq 0$$ therefor it is a strictly convex function.