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In showing that if $f,g\in L^p$, then $f+g\in L^p$, one can use the fact that $$|f+g|^p\leq 2^p\left(|f|^p + |g|^p\right).$$

The result I'd like help in proving is this:

Given that $1\lt p \lt \infty$, and $\forall ~s,t\in [0,\infty)$ $$\left(s+t\right)^p \leq 2^p\left(s^p+t^p\right).$$

Thank you.

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Do you mean $(s+t)^p\leq 2^p(s^p+t^p)$? – Paul Nov 16 '11 at 8:29
Yes Paul. I'll edit to that effect. Thanks. – Colin Nov 16 '11 at 8:35
To invert a common saying, we mathematicians often lose sight of the trees for the forest. Some very basic examples would have convinced you that $ (s+t)^p\leq 2^p(s^p+t^p)$ is indeed true, but incredibly poor as a bound. For example with $s=3$, $t=4$ it gives $49\leq 100$. The situation begs for us to try $ (s+t)^p\leq 2^{p-1}(s^p+t^p)$ which -as other users have pointed out- is indeed provable. For the case above it gives you the much stronger $49\leq 50$. – Naima65 Aug 13 at 12:59

3 Answers 3

up vote 8 down vote accepted

If I am not mistaken more is actually true. Since for $p>1$ the function $$f(x)=x^p$$ is convex, then by the midpoint inequality one has $$f\left(\frac{x+y}{2}\right)\leq \frac{f(x)+f(y)}{2}.$$ This translates immediately in $$(x+y)^p\leq 2^{p-1}(x^p+y^p),$$ which in turn imply your thesis. The case left is then $p=1$ but in this case the result is trivial.

Note that one can say more even in the case $0\leq p<1$. Indeed $\forall t\in [0,+\infty)$ one gets $$(1+t)^p\leq 1+t^p.$$ To prove the result define $$k(t)=(1+t)^p-t^p.$$ Then $\forall t\in (0,+\infty)$, we have $$k'(t)=p[(t+1)^{p-1}-t^{p-1}]<0,$$ and $$\lim_{t\to 0}\:k(t)=1,$$ therefore the claim follows. Now set $$t=\frac{x}{y}$$ to conclude $$(x+y)^p\leq x^p+y^p.$$

EDIT: According to Martin Sleziak's comment, this imply in turn that the function $$f(x)=x^p$$ is subadditive for $p<1$.

To collect all the informations all at once, one usually says that $\forall p\in [0,+\infty)$ it is true that $$(x+y)^p\leq \max(1,2^{p-1})(x^p+y^p).$$

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Nice. Perhaps it's worth mentioning that what you proved in the second part is that the function $f(x)=x^p$ is subadditive for $p<1$. – Martin Sleziak Nov 16 '11 at 12:06

The stronger (and optimal) result is that, for every $\color{red}{p\geqslant1}$ and nonnegative $s$ and $t$, $$ \color{red}{(s+t)^p\leqslant2^{p-1}(s^p+t^p)}. $$ A simple method of proof is to note that one wants to show that $c\left(\frac12(s+t)\right)\leqslant\frac12\left(c(s)+c(t)\right)$, where $c$ denotes the function defined on $x\geqslant0$ by $c(x)=x^p$, that this inequality is the midpoint property for a convex function, and that $c$ is indeed convex for every $p\geqslant1$.

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So you want to prove: Given that $ 1 < p < \infty $ and $ \forall s,t \in [0,\infty) $

$(s+t)^p \le 2^p(s^p+t^p) $


$ (s+t)^p \le (2 \max(s,t))^p \le 2^p (s^p+t^p) $

and you're done.



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