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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?

Axion004
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2 Answers2

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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 $2^{p-1} \leq 2^p$.

N. S.
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  • I see, the inequality follows from what is known as Hölder's inequality for sums (called the discrete Hölder inequality) where $$\sum_{k=1}^n |a_k,b_k| \le \biggl( \sum_{k=1}^n |a_k|^p \biggr)^{\frac{1}{p}} \biggl( \sum_{k=1}^n |b_k|^q \biggr)^{\frac{1}{q}} \text{ for all }(a_1,\ldots,a_n),(b_1,\ldots,b_n)\in\mathbb{R}^n\text{ or }\mathbb{C}^n$$ – Axion004 Mar 29 '21 at 15:18
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Alternatively, observe that $a \ge 1, b \ge 1$ and you prove: $f(x)=x^t+(1-x)^t \ge \left(\dfrac{1}{2}\right)^t, 0 < x < 1$. Take $f'(x) = tx^{t-1}-t(1-x)^{t-1}=0\iff x=\dfrac{1}{2}$, and $f''(\frac{1}{2})> 0$, hence $f_{\text{min}} = f(\frac{1}{2}) = \dfrac{1}{2^{t-1}}\ge \dfrac{1}{2^t}$.