Is $L^p$ linear for $0<p<1$?

Question

The following is an exercise from Bruckner's Real Analysis:

Show that for all $0 <p< \infty$ the collections $L^p$ of measurable functions defined on a measure space $(X, \mathcal{M},μ)$ such that $ \int_X |f|^p d\mu < \infty$ are linear spaces. [Hint : Use the inequality $(a + b)^p ≤ 2^p(a^p + b^p)$.]

The spaces is closed for all $p$ under scalar multiplication but for closed under sum we use the hint : The case $x=0$ is obvious so letting $a>0$ and considering $x=a/b$, we have $f(x)=(x + 1)^p - 2^p(x^p + 1) \le 0$ holds for all $0\le x$ when $1 \le p$ but doesn't hold for $0<p<1$ always. So does really the claim fails for $0<p<1$ or I make mistakes?

So the question is about validity of $f(x)=(x + 1)^p - 2^p(x^p + 1) \le 0$ for $0<p<1$.

Answer 1

Hint:

• For $0<p<1$: you have the inequality $(a+b)^p\leq a^p+b^p$ for all $a,b\geq0$. This can be proved by looking at $\phi(x)=(1+x)^p-x^p$. Using simple differential calculus you can show that $\phi(x)$ is monotone decreasing.
  Then, for $f,g\in L_p$ you have that $$\int|f+c g|^p\,d\mu\leq \int|f|^p\,d\mu + c^p\int|g|^p\,d\mu$$

Observation: unlike the case $p\geq1$, On spaces $L_p$ with $0<p<1$, $\Big(\int|f|^p\,d\mu\Big)^{1/p}$ is not a norm. However $$ d(f,g):=\int|f-g|^p\,d\mu$$ does define a complete metric in such spaces.

• For $1\leq p$, if you know Jensen's inequality you can see that $x\mapsto x^p$ is convex and so, for $a,b\geq0$ $$(a+b)^p=2^p\Big(\frac{a+b}{2}\Big)^p\leq 2^p\frac{a^p+b^p}{2}=2^{p-1}(a^p+b^p)$$

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There are other simpler ways to del with the inequality in your hint. For example, for $a,b\geq0$, you have $a+b\leq 2\max(a,b)$. As $x\mapsto x^p$ is increasing $$(a+b)^p\leq 2^p\Big(\max(a,b)\Big)^p\leq 2^p(a^p+b^p)$$

Answer 2

Note:

1. For $0<p<+\infty$ , note that $x \mapsto x^p$ is an increasing function for $x \geq 0$.

2. For any $a, b> 0 $, $a+b \leq 2 \max(a,b)$.

Combining 1 and 2, we have $$ (a+b)^p \leq 2^p \max(a,b)^p \leq 2^p (a^p +b^p) \tag{1}$$

So we have proved that $(1)$ is true for any $0<p<+\infty$ .

Now, using $(1)$, for any $0<p<+\infty$ and $f, g \in L^p$, we have $$\int |f +g|^p d \mu \leq \int (|f| + |g|)^p d \mu \leq \int 2^p(|f|^p + |g|^p) d \mu \leq 2^p \left ( \int |f|^p d \mu + \int |g|^p d \mu \right ) < \infty$$ So $f+g \in L^p$.

Remark: Answering the last part of your question: "So the question is about validity of $f(x)=(x + 1)^p - 2^p(x^p + 1) \le 0$ for $0<p<1$".

From $(1)$, we have that, for any $0 <p < +\infty$, if $x >0$ then $$ (x + 1)^p \leq 2^p(x^p + 1) $$ So $f(x)=(x + 1)^p - 2^p(x^p + 1) \le 0$ for $0<p<1$.