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Let $\left\|x \right\| = \sum_{i=1}^{i=n}\left|x^i\right|$ and $d\left(x\right)=\sum_{i=1}^{i=n}x^i\ln x^i$ where $x\in R^n $ and $ \sum_{i=1}^{i=n}x^i=1$

How to prove: For all $x, x'$, $$\left| d\left(x\right)-d(x') \right|\leq \frac{1}{2}\left\|x-x' \right\|$$

I would appreciate it if you could give me any information about this. I had looked so hard but got nothing.


By the way, the above question is equivalent to the question below. Let $\left\|x \right\| = \sum_{i=1}^{i=n}\left|x^i\right|$ and $d\left(x\right)=\sum_{i=1}^{i=n}x^i\ln x^i$, where $x\in \left\{R^n \mid x\ge0,\sum_{i=1}^{i=n}x^i=1\right\}$ The question is how to prove $d_1\left(x\right)\ge \frac{1}{2}\left\|x-x_0\right\|$ where $x_0= \operatorname{argmin}_x \left\{d_1\left(x\right)\right\}$. Actually $x_0^i=\frac{1}{2},i=1,2,...,n$

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Cross posted on Mathoverflow… – Dominic Michaelis Apr 3 '13 at 8:59
Does someone who conducts math career likes a spy or detective´╝Ť-) I had already deleted it. A math researcher told me that if there is no answer after one or two weeks, I could post to overflow. – Vivian Apr 3 '13 at 10:03
Why did you delete it then? Did you get your hint? If not, you should undelete it. – abatkai Apr 3 '13 at 10:22
I happliy found the answer from a scribe of Ofer Dekel and Albert Yu.… It gave a method to prove it. – Vivian Apr 3 '13 at 11:20
@abatkai Hi. The above link gives a hint. It shows a way to prove. – Vivian Apr 3 '13 at 11:25

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