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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 $f(x)=\frac{1}{2}\langle Ax,x\rangle - \langle b,x\rangle$

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    $\begingroup$ This is obviously not true for any function. What are your assumptions about $f$? $\endgroup$ – Adam Latosiński May 4 at 15:23
  • $\begingroup$ You mean if $f$ satisfies the defintion in the linked question? $$\begin{align*} f(y)\geq f(x) + \langle \nabla f(x),y-x\rangle + \frac{m}{2}\|y-x\|_2^2, \end{align*}$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 4 at 15:26
  • $\begingroup$ @AdamLatosiński I'm sorry, I've edited the question. $\endgroup$ – Ilan Aizelman WS May 4 at 15:32
  • $\begingroup$ @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 $\phi$ $\endgroup$ – Ilan Aizelman WS May 4 at 15:33
  • $\begingroup$ What is the domain of $x$ and $y$? $f$ is the solution of the integral solution? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 4 at 15:37
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$f$ 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)<tf(x_1)+(1-t)f(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.

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