Two Endpoint Triangle Inequalities
At the two endpoints \(p=1\) and \(p=\infty\), the triangle inequality has the same form, but its proof uses different features of the norm. For \(L^1\), the pointwise inequality for absolute values can be integrated. For \(L^\infty\), one combines almost-everywhere bounds, taking care that the exceptional null sets for two functions may differ. Neither argument requires the measure space to have finite measure.
Let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, complete, or sigma-finite. Functions in this tutorial are measurable real-valued or complex-valued functions, considered up to equality \(\mu\)-almost everywhere. The spaces \(L^1(\mu)\) and \(L^\infty(\mu)\) consist, respectively, of equivalence classes of functions with finite integral of the absolute value and of essentially bounded functions. Their norms are
The essential supremum is the least almost-everywhere upper bound: for a nonnegative measurable function \(h\), it is \(\inf\{C\ge 0:h\le C\ \mu\text{-a.e.}\}\), with value \(+\infty\) if no finite such bound exists. In particular, if \(\|f\|_\infty=a<\infty\), then for every \(\varepsilon>0\), \(|f|\le a+\varepsilon\) almost everywhere. The norm is a norm on the equivalence classes, not on pointwise representatives: changing a function on a null set does not change either norm.
Proof. First suppose \(f,g\in L^1(\mu)\). For almost every \(x\), the scalar triangle inequality gives \(|f(x)+g(x)|\le |f(x)|+|g(x)|\). The right side is integrable because \(f,g\in L^1(\mu)\), and monotonicity and additivity of the integral of nonnegative integrable functions give
Thus \(f+g\in L^1(\mu)\), and the same inequality is the asserted norm bound.
Now suppose \(f,g\in L^\infty(\mu)\), and put \(a=\|f\|_\infty\), \(b=\|g\|_\infty\), both finite. Fix \(\varepsilon>0\). By the defining property of essential supremum, there are measurable null sets \(N_f,N_g\) such that \(|f|\le a+\varepsilon/2\) outside \(N_f\), and \(|g|\le b+\varepsilon/2\) outside \(N_g\). The union \(N_f\cup N_g\) is null. Outside that union, the scalar triangle inequality yields \(|f+g|\le a+b+\varepsilon\). Therefore \(\|f+g\|_\infty\le a+b+\varepsilon\) for every \(\varepsilon>0\). If this norm were greater than \(a+b\), choosing \(\varepsilon\) smaller than their positive difference would contradict that bound. Hence \(\|f+g\|_\infty\le a+b\), and \(f+g\) is essentially bounded. Finally, replacing either function by an almost-everywhere equal representative changes the sum only on a union of null sets, so the conclusions are statements about equivalence classes. \(\square\)
Worked Comparisons
Worked Example: A Finite Counting Measure Space
Let \(X=\{1,2,3\}\), let \(\mathcal{M}\) be its power set, and use counting measure \(\mu\), for which every point has measure \(1\). Set \(f=(1,-2,0)\) and \(g=(2,1,-1)\), with coordinates listed in the order \(1,2,3\). Compute both endpoint inequalities.
For counting measure, the \(L^1\) norm is the sum of the absolute values. Thus \(\|f\|_1=|1|+|-2|+|0|=1+2+0=3\), and \(\|g\|_1=|2|+|1|+|-1|=2+1+1=4\). Their sum is \(f+g=(1+2,-2+1,0-1)=(3,-1,-1)\), so \(\|f+g\|_1=|3|+|-1|+|-1|=3+1+1=5\). The inequality reads \(5\le3+4=7\).
For \(L^\infty\) on this finite space, the essential supremum is the maximum over the three points, since each point has positive measure. Therefore \(\|f\|_\infty=\max\{1,2,0\}=2\), \(\|g\|_\infty=\max\{2,1,1\}=2\), and \(\|f+g\|_\infty=\max\{3,1,1\}=3\). Here the inequality reads \(3\le2+2=4\). Check: the coordinatewise sums are \(3,-1,-1\), whose absolute values sum to \(5\) and whose maximum is \(3\), matching both computed norms.
Worked Example: Complementary Functions on the Unit Interval
Use Lebesgue measure \(m\) on \(X=[0,1]\), with its Lebesgue sigma-algebra. Let \(f(x)=x\) and \(g(x)=1-x\). Compute the \(L^1\) and \(L^\infty\) norms of each function and their sum.
Since both functions are nonnegative on \([0,1]\), their \(L^1\) norms are their integrals. Direct integration gives \(\|f\|_1=\int_0^1x\,dx=[x^2/2]_0^1=1/2\), and \(\|g\|_1=\int_0^1(1-x)\,dx=[x-x^2/2]_0^1=1-1/2=1/2\). Their pointwise sum is \(f(x)+g(x)=x+(1-x)=1\), so \(\|f+g\|_1=\int_0^1 1\,dx=1\). The \(L^1\) inequality is an equality: \(1=1/2+1/2\).
The essential supremum of \(f\) is \(1\), even though \(f(x)<1\) for every \(x<1\): for every \(0<\varepsilon<1\), the set where \(x>1-\varepsilon\) has measure \(\varepsilon>0\), so no number below \(1\) is an almost-everywhere upper bound. The same argument, using the interval \(0\le x<\varepsilon\), gives \(\|g\|_\infty=1\). Since \(f+g=1\) everywhere, \(\|f+g\|_\infty=1\), and the \(L^\infty\) inequality is \(1\le1+1=2\). Check: the sum is exactly the constant function \(1\), whose integral and essential supremum on a set of measure \(1\) are both \(1\).
Worked Example: Disjoint Supports on the Real Line
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. Define \(f=\chi_{[0,1]}\) and \(g=2\chi_{[2,3]}\). The intervals are disjoint and each has measure \(1\). Compute both endpoint inequalities.
For \(L^1\), \(\|f\|_1=\int_{\mathbb{R}}\chi_{[0,1]}\,dm=m([0,1])=1\). Also, \(\|g\|_1=\int_{\mathbb{R}}2\chi_{[2,3]}\,dm=2m([2,3])=2\). On \([0,1]\), \(f+g=1\); on \([2,3]\), \(f+g=2\); outside their union it is \(0\). Hence \(\|f+g\|_1=1\cdot m([0,1])+2\cdot m([2,3])=1+2=3\), and the bound is \(3\le1+2=3\).
For \(L^\infty\), the values of \(|f|\) are at most \(1\), and \(|f|=1\) on a set of measure \(1\), so \(\|f\|_\infty=1\). The values of \(|g|\) are at most \(2\), and \(|g|=2\) on a set of measure \(1\), so \(\|g\|_\infty=2\). The sum takes value \(1\) on \([0,1]\), value \(2\) on \([2,3]\), and \(0\) elsewhere; therefore \(\|f+g\|_\infty=2\). The inequality is \(2\le1+2=3\). Check: the integral adds the contributions from both disjoint intervals, while the essential supremum is the larger of the two values, \(2\).
Why the Two Arguments Differ
The \(L^1\) estimate uses the integral to aggregate values over the whole space. It applies whether or not the supports overlap and whether or not the measure space has finite measure, provided the two functions are individually in \(L^1(\mu)\). The \(L^\infty\) estimate instead controls values outside null sets. Its proof must account for both exceptional sets: outside their union, the two bounds hold at once. A union of two null sets is still null, so this step does not require completeness of the measure space.
A frequent error is to replace an essential supremum by a pointwise supremum. For a concrete correction, let \(X=[0,1]\) with Lebesgue measure and \(f=\chi_{\mathbb{Q}\cap[0,1]}\). The rationals in \([0,1]\) are countable and hence have Lebesgue measure zero. Pointwise, \(\sup_{x\in[0,1]}|f(x)|=1\), but \(f=0\) almost everywhere. Thus \(\|f\|_\infty=0\), not \(1\), and \(\|f\|_1=0\).
Wrong calculation: “The largest value of \(f\) is \(1\), so \(\|f\|_\infty=1\).” This uses the pointwise supremum and ignores that the set where \(f=1\) is null. Correction: for every \(C>0\), the inequality \(|f|\le C\) holds almost everywhere, because it fails only on the rationals. The least such almost-everywhere bound is \(0\). In particular, the zero function and \(f\) represent the same element of both \(L^1(\mu)\) and \(L^\infty(\mu)\).
Check Your Understanding
For each question, distinguish a pointwise statement from an almost-everywhere one where needed.
- On a finite space with counting measure, let \(f=(2,0)\) and \(g=(-1,3)\). Compute \(\|f\|_1,\|g\|_1,\|f+g\|_1\) and the corresponding \(L^\infty\) norms.
- Explain why the \(L^1\) proof guarantees \(f+g\in L^1(\mu)\), rather than merely giving a formal inequality, when \(f,g\in L^1(\mu)\).
- In the \(L^\infty\) proof, why is it necessary to combine the exceptional null sets for \(f\) and \(g\)?
- On \([0,1]\) with Lebesgue measure, find the pointwise supremum and essential supremum of \(\chi_{\mathbb{Q}\cap[0,1]}\).
- For \(f=\chi_{[0,1]}\) and \(g=\chi_{[1,2]}\) on \(\mathbb{R}\) with Lebesgue measure, compute both \(L^1\) norms and the \(L^1\) norm of \(f+g\). Does the shared endpoint affect the result?