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

  • 1
    $\begingroup$ 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

$f$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.