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Let $n\in \mathbb{N}$ and $f:\mathbb{R}\rightarrow\mathbb{R}$

$$f(\sum\limits_{k=1}^n \lambda_k x_k) \leq \sum\limits_{k=1}^n \lambda_k f(x_k)$$

where $\sum\limits_{k=1}^n \lambda_k = 1$ and arbitrary $x_k \in \mathbb{R}$ and fixed $\lambda_k \in \mathbb{R} $ with the fixed sum of 1.

Do I understand that correctly, that $\sum\limits_{k=1}^n \lambda_k = 1$ for any $n$ or does it hold only for a specific $n$ and not necessarily $\sum\limits_{k=1}^{n+1} \lambda_k = 1$ for example?

Furthermore, I'd use induction to prove it, the start is trivial so I'll only show the step from n to n+1.

$$f(\sum\limits_{k=1}^{n+1} \lambda_k x_k)=f(\sum\limits_{k=1}^n \lambda_k x_k+\lambda_{n+1}x_{n+1})$$ but here's the problem because in order to use the hypothesis, I'd have to look at $f(\sum\limits_{k=1}^n \lambda_k x_k)$ alone, and I don't know how to "split it". I know I got to use the fact, that the function is convex, so it's increasing, continous and I got the defintion of a convex function. Can somebody give me a hint on what tool to use?

pshmath0
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Jonathan
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2 Answers2

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You need to use both convexity and the induction hypothesis.

$$\begin{align}f(\sum\limits_{k=1}^{n+1} \lambda_k x_k)&=f(\sum\limits_{k=1}^n \lambda_k x_k+\lambda_{n+1}x_{n+1})\\ &=f\left[\left(\sum_{j=1}^n\lambda_j\right) \sum_{k=1}^n \frac{\lambda_k}{\sum_{j=1}^n\lambda_j}x_k + \lambda_{n+1}x_{n+1}\right]\\ &= f\left((1-\lambda_{n+1})\sum_{k=1}^n \frac{\lambda_k}{\sum_{j=1}^n\lambda_j}x_k + \lambda_{n+1}x_{n+1} \right) \\ &\leq (1-\lambda_{n+1})f\left(\sum_{k=1}^n \frac{\lambda_k}{\sum_{j=1}^n\lambda_j}x_k \right) + \lambda_{n+1}f(x_{n+1}) \\ &\leq \sum_{j=1}^n\lambda_j \sum_{k=1}^n \frac{\lambda_k}{\sum_{j=1}^n\lambda_j}f(x_k) + \lambda_{n+1}f(x_{n+1}) \\ &= \sum_{k=1}^{n+1}\lambda_k f(x_k) \end{align}$$

When $\sum_{j=1}^n\lambda_j=0$, $\lambda_1=\ldots=\lambda_n=0$ and $\lambda_{n+1}=1$, so the clain holds.

Gabriel Romon
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  • I see where the convexity comes in, but what happens in the step ( second, third line ) where you go from $\sum_{j=1}^n\lambda_j$ to $(1-\lambda_{n+1})$? Isn't $\sum_{j=1}^n\lambda_j$ equal to 1? Would $\lambda_{n+1}$ then be equal to zero? – Jonathan Mar 20 '17 at 16:38
  • @JonathanZ $n$ is the variable upon which you're performing induction. During the inductive step, every $n$ should be replaced with $n+1$: you have to prove that for any $\lambda_1,\ldots,\lambda_{n+1}$ such that $\sum_{k=1}^{n+1} \lambda_k =1$, $$f(\sum\limits_{k=1}^{n+1} \lambda_k x_k) \leq \sum\limits_{k=1}^{n+1} \lambda_k f(x_k)$$ – Gabriel Romon Mar 20 '17 at 16:44
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    I used the induction hypothesis on this specific set: $$\frac{\lambda_1}{\sum_{j=1}^n\lambda_j},\ldots,\frac{\lambda_n}{\sum_{j=1}^n\lambda_j}$$ These are $n$ non-negative real numbers that sum to $1$. – Gabriel Romon Mar 20 '17 at 16:46
  • Wow, that took me a while, excuse me! I had some trouble seeing how you applied the hypothesis but it's clear now – Jonathan Mar 20 '17 at 17:04
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When you say "fixed $\lambda_k$" it reminds me of this: A function $f$ is called midpoint convex if $$ f\left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2} $$ for all $x,y$. In general (lacking the assumption of continuity), a midpoint convex function need not be convex.

GEdgar
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