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Let $x,y \in \mathbb{R}^n$, $f: \mathbb{R}^n \to \mathbb{R}^n$ a continuous map.

(Per comment) Let $A$ be an symmetric, positive definite matrix.

I wish to prove that:

$(A f(x) - A f(y))^T(x - y) \geq \lambda_{min}(A)(f(x) - f(y))^T(x - y) $

$\lambda_{min}$ being the smallest eigenvalue of $A$. Can anyone see what principle is used such that the inequality is true?

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    I think I am misinterpreting the question... if $A=\begin{bmatrix}1 & 0\\0 & 2\end{bmatrix}$ and $x=f(x)=(0,0)$ and $y=(2,-1)$ and $f(y)=(1,1)$ doesn't the left hand side become 0 but the right hand side stay positive? – 2017-01-12
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    @alphacapture There maybe more assumptions such that this condition is true. For example $(f(x) - f(y))^T(x-y) \geq 0, \forall x,y$. – 2017-01-12
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    What is the context? – 2017-01-12
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    This question doesn't make any sense. If $A$ is indeed arbitrary, it may possess non-real eigenvalues. What does $\lambda_\min(A)$ mean then? – 2017-01-12
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    @user1551 Nice! You just answered the restriction part. Let $A$ be symmetric positive definite instead. See, it does make sense. – 2017-01-12
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    In that case the problem statement does make sense, but it definitely doesn't hold when $f(x)=-x$ and $A$ is not a multiple of $I$. – 2017-01-12
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    @user1551 Assume further that $(f(x) - f(y))^T(x-y) \geq 0, \forall x, y \in \mathbb{R}^n$ – 2017-01-12

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The question simplifies to:

suppose $A$ is positive definite and $x^Ty \geq 0$, show that $(Ax)^T y \geq \lambda_{min}(A) x^T y$

Writing $A = \sum_i \lambda_i v_i v_i^T$, we need to show $$\sum_i \lambda_i x^T v_i v_i^T y \geq \lambda_{min}(A) x^T y$$

This is not necessarily true. Take $n=2$, $\lambda_1 = 1$, $\lambda_2 = 0.5$, $$ x = \begin{pmatrix}1\\0\end{pmatrix}, y = \begin{pmatrix}0\\1\end{pmatrix}, v_1 = \begin{pmatrix}\sqrt{2}/2 \\ -\sqrt{2}/2\end{pmatrix}, v_2 = \begin{pmatrix}\sqrt{2}/2 \\ \sqrt{2}/2\end{pmatrix}. $$ The statement is now $-0.5 \geq 0$, which is clearly not true. Returning to your original formulation, the function $f(x) = \begin{pmatrix}x_2 \\ x_1\end{pmatrix}$, the points $x=\begin{pmatrix}1 \\ 1\end{pmatrix}$, $y=\begin{pmatrix}1 \\ 0\end{pmatrix}$, and $A=\begin{pmatrix}0.75 & 0.75 \\ -0.25 & -0.25\end{pmatrix}$ are a counterexample.

For the statement to hold, you need $x^T v_i v_i^T y \geq 0$. In that case: $$\sum_i \lambda_i x^T v_i v_i^T y \geq \sum_i \lambda_{min}(A) x^T v_i v_i^T y = \lambda_{min}(A) x^T \left(\sum_i v_i v_i^T \right) y = \lambda_{min}(A) x^Ty$$

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    What if $A$ was positive semidefinite. – 2017-01-12
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    Positive definite implies positive semidefinite, so the same counterexample works. You could take $\lambda=0$ if you want a positive semidefinite $A$ that is not positive definite. – 2017-01-12