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Inequality and maximum

  1. Dec 9, 2009 #1
    Hi guys

    Today my lecturer wrote on the blackboard

    \max \left\{ {\alpha f(x) + (1 - \alpha )f(y)\,\,,\,\,\alpha g(x) + (1 - \alpha )g(y)} \right\}\,\,\,\, \le \,\,\,\alpha \max \left\{ {f(x)\,\,,\,\,g(x)} \right\} + (1 - \alpha )\max \left\{ {f(y)\,\,,\,\,g(y)} \right\},

    where x, y are variables in all R, and alpha is a constant in [0;1]. I must admit, I cannot quite see why this inequality holds. Are there some rules about the maximum that is being used here?
  2. jcsd
  3. Dec 9, 2009 #2
    Ok, starting with

    [tex]\max \left\{ \alpha f(x) + (1 - \alpha )f(y),\alpha g(x) + (1 - \alpha )g(y) \right\}[/tex]

    Obviously, [tex]f(x) \le \max \left\{f(x), g(x)\right\} = x_{max}[/tex], and similar for [tex]g(x)[/tex], so we can safely use this to get something which is greater or equal:

    [tex]\max \left\{ \alpha f(x) + (1 - \alpha )f(y),\alpha g(x) + (1 - \alpha )g(y) \right\} \le \max \left\{ \alpha x_{max} + (1 - \alpha )f(y),\alpha x_{max} + (1 - \alpha )g(y) \right\}[/tex]

    Again, [tex]f(y) \le \max \left\{f(y), g(y)\right\} = y_{max}[/tex], and similar for [tex]g(y)[/tex], and using that we get

    [tex]\max \left\{ \alpha x_{max} + (1 - \alpha )f(y),\alpha x_{max} + (1 - \alpha )g(y) \right\} \le \max \left\{ \alpha x_{max} + (1 - \alpha )y_{max},\alpha x_{max} + (1 - \alpha )y_{max} \right\}[/tex]

    Now, for [tex]a \ge 0[/tex] we have that [tex]\max \left \{ ab, ac \right \} = a \max \left \{b, c \right \}[/tex] and [tex]\max \left \{ a + b, a + c \right \} = a + \max \left \{b, c \right \}[/tex]. Assuming [tex]\alpha \in [0,1][/tex], we can use this to get

    [tex]\max \left\{ \alpha x_{max} + (1 - \alpha )y_{max},\alpha x_{max} + (1 - \alpha )y_{max} \right\} = \alpha x_{max} + (1 - \alpha )\max \left\{y_{max}, y_{max} \right\} = \alpha x_{max} + (1 - \alpha )y_{max}[/tex].

    Inserting the expressions for [tex]x_{max}[/tex] and [tex]y_{max}[/tex] gets you to the right hand side you had, which thus is equal or greater than the left hand side.
  4. Dec 10, 2009 #3
    Thank you. It is very kind of you to help.
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