A Why the triangle inequality is greater than the 2 max{f(x),g(x)}

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The discussion centers on a proof from Sheldon Axler's book on measure theory, specifically regarding the properties of the space L^p(μ). It examines the triangle inequality and how it leads to the conclusion that the norm of the sum of two functions is bounded by a function of their individual norms. The key point is that if one function dominates the other in absolute value, the sum can be simplified to show that it adheres to the inequality involving the maximum of the two functions. Understanding these inequalities is crucial for proving that L^p(μ) is a vector space with standard operations. This foundational concept is essential for further exploration in measure theory.
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I am working on proving the L^p of measureable space (X,S,u) is a vector space. I am lost on why the the triangle intequalty is greater than the 2 max{f(x),g(x)} for a fix x in X.
I am reading Sheldon's Axler Book on Measure theory. He is proving that ##L^p(\mu)## is a Vector space over ##\mathbb{R}.## He claims that if ##\|f+g\|_{p}^p\leq 2^p(\|f\|_{p}^p+\|g\|_{p}^{p})## and nonzero homogenity holds true, then ##L^p{\mu}## is true with the standard addition and scalar multiplication. He starts with the following assumption:

Suppose that ##f,g\in L^p(\mu)## are arbitrary. Then if ##x\in X## is an arbitrary fix element of ##X,## then
##\begin{align*} |f(x)+g(x)|^p&\leq_{\text{triangle inequality}} (|f(x)|+|g(x)|)^p\\ &\leq_{\text{why?}} (2\max{|f(x)|,|g(x)|})^p\\ &\leq2^p(|f(x)|^p+|g(x)|^p)\end{align*}##

If you can explain whys in this proof then I will be able to understand the proof.

Thanks,

Carter Barker
 
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Because if, e.g., |f|>=|g|, then |f|+|g|<=|f|+|f|=2|f|
 
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