Let $u(t):=t\log t$. This function is convex on $[0,+\infty)$ and $u(0)=0$. How can I prove that$$u(t+h)-u(t)\geq u(h)\quad\forall t,h\in[0,+\infty)?$$ Since $u(0)=0$ using convexity we know that $u(st)\leq s\,u(t)$, where $s\in[0,1]$ but i get confused with the inequalities.
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1$-u$ is concave, and therefore subadditive, see for example https://math.stackexchange.com/a/2676802/42969. – Martin R Jun 04 '21 at 06:46
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1Does this answer your question? If $f$ is a concave function and $f(0) \geq 0$ then $f(a+b) \leq f(a) + f(b)$ for all $a,b >0$ – Martin R Jun 04 '21 at 08:21
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It is one line (\log is monotone):
$$u(t+h)=(t+h)\log(t+h)=t\log(t+h)+h\log(t+h)\ge t\log(t)+h\log(h)=u(t)+u(h).$$
crankk
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No, $x\varphi(x)$ with $\varphi\nearrow$ will work the same. For instance $x\tanh(x)$ is not convex, but the result stands. – zwim Jun 03 '21 at 20:01
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Note, that
$$\begin{array}{rcl} u(t)+u(h)&=&u((t+h)\dfrac{t}{(t+h)})+u((t+h)\dfrac{h}{(t+h)}) \\ &\leq& \dfrac{t}{(t+h)}\cdot u(t+h)+\dfrac{h}{(t+h)}\cdot u(t+h)=u(t+h) \quad \forall t,h\in[0,+\infty) \end{array}$$
motionart
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So this would work also for any function which is convex on $(0,+\infty)$ without using the logarithm of this $u$! Interesting – Mathemachicken Jun 03 '21 at 20:00
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