The Usual \(L^p\) Functional
For \(p\ge1\), the \(p\)th-root integral is a norm on \(L^p(\mu)\), with functions identified when they agree almost everywhere. The range \(0<p<1\) behaves differently: even on a space with only two points, the same formula can assign a larger value to a sum than the sum of the assigned values. The failure is not a matter of a subtle limiting example. It is visible in a direct finite calculation.
Let \((X,\mathcal{M},\mu)\) be any measure space, with no assumption that \(\mu\) is finite or sigma-finite. Let \(f:X\to\mathbb{R}\) or \(f:X\to\mathbb{C}\) be measurable. For \(0<p<1\), define \(L^p(\mu)\) to consist of equivalence classes, under equality \(\mu\)-a.e., of such functions for which \(\int |f|^p\,d\mu<\infty\). The usual \(p\)th-root functional is
The integral and the resulting functional are unchanged by altering \(f\) on a null set. We will see that this functional is homogeneous and positive definite on these equivalence classes, but does not satisfy the norm triangle inequality. In particular, calling it a norm for \(0<p<1\) would be incorrect.
Worked Example: Two Disjoint Points Give a Failure
Take \(X=\{1,2\}\), \(\mathcal{M}\) the collection of all subsets of \(X\), and \(\mu\) counting measure. Thus \(\mu(\{1\})=\mu(\{2\})=1\), and integration is summation over the two points. Let \(f(1)=1,\ f(2)=0\) and \(g(1)=0,\ g(2)=1\). Both functions are real-valued and measurable. For any fixed \(p\) with \(0<p<1\),
The sum has values \((f+g)(1)=1\) and \((f+g)(2)=1\), so
Since \(0<p<1\), we have \(1/p>1\), and therefore \(2^{1/p}>2\). The triangle inequality would require \(\|f+g\|_p\le\|f\|_p+\|g\|_p=2\), which is false. For a fully numerical instance, take \(p=\tfrac12\): then \(\|f+g\|_{1/2}=2^2=4\), whereas \(\|f\|_{1/2}+\|g\|_{1/2}=1+1=2\). Check: substituting the two values of \(f+g\) into the defining formula gives \((1^{1/2}+1^{1/2})^{2}=(1+1)^2=4\), confirming the violation.
The Inequality That Replaces the Triangle Inequality
The usual triangle inequality fails, but a related inequality holds. Its source is a scalar inequality for powers below one.
Proof. If \(a=0\), then \((a+b)^p=b^p=0^p+b^p\); if \(b=0\), equality also holds. Now suppose \(a>0\) and \(b>0\), and set \(t=b/a>0\). It is enough to prove \((1+t)^p\le1+t^p\), because multiplying this inequality by \(a^p\) gives the claimed result. Define \(h(t)=1+t^p-(1+t)^p\) for \(t>0\). Its derivative is \(h'(t)=p\bigl(t^{p-1}-(1+t)^{p-1}\bigr)\). Since \(p-1<0\) and \(0<t<1+t\), raising these positive numbers to the negative power \(p-1\) reverses their order, so \(t^{p-1}>(1+t)^{p-1}\). Hence \(h'(t)>0\). Also \(h(t)\to0\) as \(t\downarrow0\). Thus \(h(t)\ge0\) for \(t>0\), proving the inequality in all cases. \(\square\)
Proof. At each \(x\in X\), the ordinary triangle inequality for absolute values gives \(|f(x)+g(x)|\le |f(x)|+|g(x)|\). The function \(s\mapsto s^p\) is increasing for \(s\ge0\), so the lemma yields \(|f(x)+g(x)|^p\le (|f(x)|+|g(x)|)^p\le |f(x)|^p+|g(x)|^p\). Integrating this pointwise inequality and using monotonicity and additivity of the nonnegative Lebesgue integral gives \(\int |f+g|^p\,d\mu\le\int |f|^p\,d\mu+\int |g|^p\,d\mu<\infty\). Thus \(f+g\in L^p(\mu)\), and taking the defining powers gives the displayed inequality. \(\square\)
The functional also has the expected homogeneity: for a scalar \(c\in\mathbb{R}\) or \(c\in\mathbb{C}\) and \(f\in L^p(\mu)\), \(\|cf\|_p=|c|\|f\|_p\). If \(c=0\), both sides are zero. If \(c\ne0\), this follows by substituting \(|cf|^p=|c|^p|f|^p\) into the integral and taking the \(1/p\) power. It is positive definite on equivalence classes as well: if \(\|f\|_p=0\), then \(\int |f|^p\,d\mu=0\). For each positive integer \(n\), the set \(E_n=\{x:|f(x)|>1/n\}\) satisfies \(\frac{1}{n^p}\mu(E_n)\le\int |f|^p\,d\mu=0\). Thus every \(E_n\) is null, and \(\{x:|f(x)|>0\}=\bigcup_{n=1}^\infty E_n\) is null. Hence \(f=0\) \(\mu\)-a.e. Conversely, a function equal to zero \(\mu\)-a.e. has integral of \(|f|^p\) equal to zero. The failed norm axiom is specifically the ordinary triangle inequality.
Worked Example: A Numerical Failure at p = 1/3
Again let \(X=\{1,2\}\) with its full sigma-algebra and counting measure, and use real-valued functions \(f(1)=1,\ f(2)=0\) and \(g(1)=0,\ g(2)=1\). Set \(p=\tfrac13\), so that \(1/p=3\). Direct substitution gives \(\|f\|_{1/3}=(1^{1/3}+0^{1/3})^3=1\) and \(\|g\|_{1/3}=(0^{1/3}+1^{1/3})^3=1\). At both points \(f+g=1\), and therefore
Check: the \(p\)-triangle inequality gives \(\|f+g\|_{1/3}^{1/3}=8^{1/3}=2\le1^{1/3}+1^{1/3}=2\), exactly as expected. The \(p\)-triangle inequality is consistent with the calculation even though the norm triangle inequality is not.
Worked Example: Failure With Overlapping Nonzero Values
On the same two-point counting space, let \(p=\tfrac12\), \(f(1)=1,\ f(2)=1\), and \(g(1)=1,\ g(2)=0\). These are real-valued measurable functions, and they are both nonzero at the first point. First,
The sum has values \((f+g)(1)=2\) and \((f+g)(2)=1\). Hence
Since \(\sqrt{2}>1\), \(3+2\sqrt{2}>5\), while \(\|f\|_{1/2}+\|g\|_{1/2}=4+1=5\). Thus the triangle inequality fails here too. Check: the \(p\)-triangle inequality reads \((3+2\sqrt2)^{1/2}\le\sqrt4+\sqrt1=3\). Both sides are nonnegative, and squaring reduces this to \(3+2\sqrt2\le9\), which holds because \(\sqrt2<3\). The weaker inequality is verified directly.
A Common Calculation Error
A frequent mistake is to apply the outer exponent as though it were \(p\), rather than \(1/p\). On \(X=\{1,2\}\) with counting measure and \(p=\tfrac12\), for the functions in the first example the correct calculation is
A wrong calculation might use \((1+1)^{1/2}=\sqrt2\). That places \(p\), rather than \(1/p=2\), outside the integral and does not follow the definition. The correction is to write the full formula first: sum the \(p\)th powers of the pointwise values, then raise that sum to \(1/p\).
A different mistake is to use Minkowski's inequality for \(0<p<1\). The standard \(L^p\) Minkowski inequality requires \(p\ge1\); its triangle conclusion cannot be carried over to this range. The two-point calculation supplies a direct test: any claimed triangle inequality for the displayed functional would imply \(4\le2\) at \(p=\tfrac12\), which is false.
What the Failure Tells Us
The example is deliberately finite: the measure space is finite, every function on it is measurable, and every integral in the calculations is a finite sum. Thus no issue involving convergence, sigma-finiteness, or changing representatives on null sets causes the failure. Both singletons have positive measure, so the values at both points genuinely contribute. More generally, the proposition shows that when \(0<p<1\), addition remains compatible with the estimate \(\|f+g\|_p^p\le\|f\|_p^p+\|g\|_p^p\), but this is not the defining triangle inequality of a norm.
The distinction matters whenever terminology or a proof depends on norm axioms. For \(0<p<1\), one must not invoke a norm triangle inequality for this functional or cite Minkowski's inequality in its usual form. Instead, use the \(p\)-triangle inequality when its hypotheses apply. The elementary two-point example is also a useful diagnostic: if a proposed general inequality is meant to hold for every measure space and every pair of functions, it must hold for this counting-measure case.
Check Your Understanding
In each question, use the definition of the functional and keep the outer exponent \(1/p\) explicit.
- On the two-point space with counting measure, take \(f=(1,0)\) and \(g=(0,1)\). Compute \(\|f\|_p,\|g\|_p,\|f+g\|_p\) for an arbitrary \(p\) with \(0<p<1\), and identify the failed norm axiom.
- For \(p=\tfrac12\), compute the functional of \((3,0)\) on the two-point counting space. Show the homogeneity relation by comparing it with the functional of \((1,0)\).
- State the \(p\)-triangle inequality for measurable \(f,g\in L^p(\mu)\), and explain why it does not assert the norm triangle inequality.
- In the calculation of \(\|f+g\|_{1/2}\) for \(f=(1,0)\), \(g=(0,1)\), what is the outer exponent? What value results?
- Why does the two-point counterexample rule out the triangle inequality for the usual \(p\)th-root functional in general, even though the functions and space are finite?