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I'm trying to find functions $f,g \in L^p(\mathbb{R})$ with $$||f+g||_p > ||f||_p+||g||_p$$ where $p \in (0,1)$. All my ideas failed so fair, any help and hints appreciated!

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Take $f$ supported on $[0,1]$ and equal to $1$, $g$ supported on $[1,2]$ and equal to $a>0$.

Then $\|f+g\|_p=(1+a^p)^{1/p}$, $\|f\|_p=1$, $\|g\|_p=a$.

We want $1+a^p > (1+a)^p$. By binomial expansion, the RHS is only $1+a/p+O(a^2)$ which is less than $1+a^p$ for $a>0$ small enough.


For who doesn't like big $O$ estimates: it suffices to prove that $$\lim_{a \to 0+}\frac{(1+a)^p-1}{a^p} < 1$$ We have by definition of derivative: $$\lim_{a \to 0+}\frac{(1+a)^p-1}{a} = p$$ and so for $p<1$: $$\lim_{a \to 0+}\frac{(1+a)^p-1}{a^p} = p \cdot \lim_{a \to 0+}a^{1-p}=0$$


The two are essentially the same, the only difference being that Taylor expansion gives the error $O(a^2)$ while the definition of derivative gives $o(a)$, which is sufficient.

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  • $\begingroup$ thank you that was elegant and simple, but one question: is it maybe easier to use Bernoulli inequality to solve the equation, i.e. $(1+a)^p \leq 1+pa < 1+a^p $ for sufficiently small $a$? $\endgroup$ – Simonsays Apr 12 '18 at 15:03
  • $\begingroup$ I don't think $(1+a)^p \leq 1+pa$, but the other way around... $\endgroup$ – punctured dusk Apr 12 '18 at 15:34
  • $\begingroup$ the 4th inequality or am I wrong? $\endgroup$ – Simonsays Apr 12 '18 at 16:15
  • $\begingroup$ $(1+a)^p$ is the larger one. Quick check: for $p$ positive integer, the inequality is obtained by forgetting terms in the binomial expansion. $\endgroup$ – punctured dusk Apr 12 '18 at 16:43

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