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Suppose $f$ and $g$ are functions from $\Bbb{R} → \Bbb{R}$. I wish to show that for any point $x_0 ∈ \Bbb{R}$, we have $D^+(f + g)(x_0) ≤ D^+f(x_0) + D^+g(x_0)$ where $D^+$ denotes the upper right-hand Dini derivative. So I tried to start with

$$ \limsup_{h \to 0^+} \frac{(f+g)(x_0 + h) - (f+g)(x_0)}{h}≤\limsup_{h \to 0^+} \frac{f(x_0 + h) - f(x_0)}{h}+\limsup_{h \to 0^+} \frac{g(x_0 + h) - g(x_0)}{h} $$

but I am not sure how to proceed.

Javi
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Homaniac
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  • Hint: $\limsup$ is subadditive https://math.stackexchange.com/questions/69391/subadditivity-of-the-limit-superior – Javi Oct 14 '18 at 10:35
  • Ah i see, Isit possible to do without considering subsequences or Epsilon Delta? – Homaniac Oct 14 '18 at 11:20
  • To prove subadditiviy? I don't think so, epsilon-delta and subsequences are part of the definition of $\limsup$. – Javi Oct 14 '18 at 11:28
  • Here I want as h goes to zero but over there is n goes to infinity or some constant though hmm – Homaniac Oct 14 '18 at 12:27
  • If you look at the statement of the first answer or the second answer, it is valid for your case too. – Javi Oct 14 '18 at 12:31
  • So I have to make use of the inequality result that was proven over there or do you mean the proof for my case is similar as in those answers? – Homaniac Oct 14 '18 at 12:36
  • You can use that result to prove your exercise. – Javi Oct 14 '18 at 12:38
  • Right so I can show $\limsup_{h \to 0^+} \frac{(f+g)(x_0 + h) }{h}≤\limsup_{h \to 0^+} \frac{f(x_0 + h) }{h}+\limsup_{h \to 0^+} \frac{g(x_0 + h) }{h}$ but how to take care of the other part hmm – Homaniac Oct 14 '18 at 12:44
  • You can use that $(f+g)(x)$ is defined to be $f(x)+g(x)$. I can write an answer if you want. – Javi Oct 14 '18 at 12:48
  • Ah if you could show it, that'll be great, thank you! :) – Homaniac Oct 14 '18 at 12:49

1 Answers1

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First use that $(f+g)(x)=f(x)+g(x)$ by definition.

$\limsup_{h \to 0^+} \frac{(f+g)(x_0 + h) - (f+g)(x_0)}{h}=\limsup_{h \to 0^+} \frac{f(x_0+h)+g(x_0 + h) - f(x_0)-g(x_0)}{h}=\limsup_{h \to 0^+}\frac{f(x_0+h)- f(x_0)+g(x_0 + h) -g(x_0)}{h}=\limsup_{h \to 0^+}\frac{f(x_0+h)- f(x_0)}{h}+\frac{g(x_0 + h) -g(x_0)}{h} (\ast)$.

Now, using subadditivity of $\limsup$

$(\ast) ≤\limsup_{h \to 0^+} \frac{f(x_0 + h) - f(x_0)}{h}+\limsup_{h \to 0^+} \frac{g(x_0 + h) - g(x_0)}{h}.$

And you're done.

Javi
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