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I proved this for convex and monotonically decreasing functions, but I didn't manage to prove it for general functions, nor did I find counterexamples.

Nicolas
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  • The question is missing context, in particular, what is your definition of a convex function and how did you prove it for decreasing convex? – A.Γ. Jul 29 '19 at 10:17
  • You recieved 3 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you. – 5xum Jul 30 '19 at 08:34

2 Answers2

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Hints:

  1. Use induction.
  2. $$\sum_{i=1}^n \lambda_i x_i = \sum_{i=1}^{n-1}\lambda_i x_i + \lambda_n x_n$$
5xum
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  • That's what I did. I used $f(\sum_{i=1}^{n+1} \lambda_ix_i) = f(\sum_{i=1}^n \lambda_i\frac{\sum_{i=1}^n\lambda_ix_i}{\sum_{i=1}^n\lambda_i}+\lambda_{n+1}x_{n+1}) \leq \sum_{i=1}^n \lambda_if(\frac{\sum_{i=1}^n\lambda_ix_i}{\sum_{i=1}^n\lambda_i})+\lambda_{n+1}x_{n+1}\leq \sum_{i=1}^n \lambda_if(\sum_{i=1}^n\lambda_ix_i)+\lambda_{n+1}x_{n+1} \leq f(\sum_{i=1}^n\lambda_ix_i)+\lambda_{n+1}x_{n+1}\leq \sum_{i=1}^n\lambda_if(x_i)+\lambda_{n+1}x_{n+1}=\sum_{i=1}^{n+1}\lambda_if(x_i)$. – Nicolas Jul 29 '19 at 10:09
  • @Nicolas Why didn't you use my hint 2 instead? – 5xum Jul 29 '19 at 10:43
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For finite sums, this may be taken as the definition of "convex". If you have some other definition, it may easily be proved, for example as hinted by 5xum.

However the question does not say that there are just finitely many $\lambda_i$. For the case of infinite sums, use the finite case together with continuity of $f$. So you need to know that Every Convex Function is Continuous

GEdgar
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