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Math Help - Proving that max{0,f(x)} is convex if f is convex

  1. #1
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    Lightbulb Proving that max{0,f(x)} is convex if f is convex

    Hi,

    I am trying to prove the above argument and have this thus far:

    Assume f(ax+(1-a)y)>=0. Then

    max{0,f(ax+(1-a)y)}=f(ax+(1-a)y)<=af(x)+(1-a)f(y) since f is convex

    I now need f(x)>=0 and f(y)>=0 so that

    af(x)+(1-a)f(y)=amax{0,f(x)}+(1-a)max(0,f(y)}

    but I don't know how to show that f(x)>=0 and f(y)>=0.

    Can anyone please help?
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  2. #2
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    Quote Originally Posted by FOLOW11 View Post
    Hi,

    I am trying to prove the above argument and have this thus far:

    Assume f(ax+(1-a)y)>=0. Then

    max{0,f(ax+(1-a)y)}=f(ax+(1-a)y)<=af(x)+(1-a)f(y) since f is convex

    I now need f(x)>=0 and f(y)>=0 so that

    af(x)+(1-a)f(y)=amax{0,f(x)}+(1-a)max(0,f(y)}

    but I don't know how to show that f(x)>=0 and f(y)>=0.

    Can anyone please help?
    You only need to use f(x)\leq max\{0,f(x)\} for all x, and apply this to the inequality that you have already got. If f(ax+(1-a)y) <0 then clearly 0=max{0,f(ax+(1-a)y)} is smaller or equal than the convex combination of two positive or 0 numbers.
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