How to show convexity for a functional

Apr 2015
62
6
Sweden
Dear all

I am considering a reflexive Banach space \(\displaystyle X\) and a Banach space \(\displaystyle Y\) together with a linear compact operator \(\displaystyle K\) such that \(\displaystyle Kx = y\).
I am then trying to show that for a given \(\displaystyle y \in Y\), the functional \(\displaystyle J : X \to \mathbb{R}\) given as:

\(\displaystyle J(x) = \left| \left| Kx - y \right| \right|_{Y}^2 + \alpha \left| \left| x \right| \right|_X^2\)

is convex. I know that this would mean:

\(\displaystyle J(\lambda x_0 + (1 - \lambda) x_1) \leq \lambda J(x_0) + (1 - \lambda) J(x_1), \: 0 < \lambda < 1 \)

My problem is that I get stuck in reaching that upper bound doing the following:

\(\displaystyle J(\lambda x_0 + (1 - \lambda) x_1) = \left| \left| K(\lambda x_0 + (1 - \lambda) x_1) - y \right| \right|_{Y}^2 + \alpha \left| \left| x \right| \right|_X^2 \)
\(\displaystyle \leq \left( \lambda \left| \left| Kx_0 - y \right| \right|_Y + (1- \lambda ) \left| \left| Kx_1 - y \right| \right|_Y + \left| \left| y \right| \right|_Y \right)^2 + \alpha \left| \left| \lambda x_0 + (1 - \lambda) x_1 \right| \right|_X^2 \)
\(\displaystyle \leq \left( \lambda \left| \left| Kx_0 - y \right| \right|_Y + (1- \lambda ) \left| \left| Kx_1 - y \right| \right|_Y + \left| \left| y \right| \right|_Y \right)^2 + \alpha \left( \lambda \left| \left| x_0 \right| \right|_X^2 + (1 - \lambda) \left| \left| x_1 \right| \right|_X^2 \right) \)

Which is where I get stuck because expanding \(\displaystyle \left( \lambda \left| \left| Kx_0 - y \right| \right|_Y + (1- \lambda ) \left| \left| Kx_1 - y \right| \right|_Y + \left| \left| y \right| \right|_Y \right)^2 \) yields to many terms to arrive at the desired result. What am I doing wrong?

Kind regards