While searching for a proof of the algebra property of Sobolev spaces ($H^s(\mathbb R^n)$ is an algebra when $s > n/2$), I found these notes. On page two the author states that if $t \ge 1$ then Hölder's inequality implies
$$((1+|x|)+(1+|y|))^t\le 2^t((1+|x|)^t + (1+|y|)^t).$$
I don't see why Hölder's inequality implies that the above inequality is true. Perhaps the author meant that for $t\ge 1$ and $a=(1+|x|),$ $b=(1+|y|)$ we obtain
$$(a+b)^t \le 2^t(a^t+b^t).$$
Something similar to Cauchy/Young's inequality might work here since
$$t = 1 \implies (a+b)\le 2(a+b),$$
$$t = 2 \implies (a+b)^2\le 4(a^2+b^2),$$
and the inequality covers the general case for $t\ge 1$. Trying
$$(a+b)^t=(a+b)(a+b)^{t-1}\le \frac{1}{2}\left(\frac{2(a+b)^2 + (a+b)^{2t}}{(a+b)^2}\right)$$
doesn't look like the correct approach. Does the inequality follow from either Young's or Hölder's inequality?
Best Answer
By Holder you have $$ (1 \cdot a+1\cdot b) \leq (a^p+b^p)^\frac{1}{p}(1+1)^\frac{1}{q} $$ Thus $$ (a+b)^p \leq 2^\frac{p}{q} (a^p+b^p) $$
Now $$ \frac{1}{p}+\frac{1}{q}=1 \Rightarrow \frac{p}{q}=p-1 $$ giving
$$ (a+b)^p \leq 2^{p-1} (a^p+b^p) $$ Now set $a=1+|x|, b=1+|y|, p=t$ and use $2^{p-1} \leq 2^p$.