The Triangle Inequality in Lp

From a Norm Inequality to a Distance

Minkowski’s inequality, proved in the previous tutorial, states that $\|f+g\|_p\leq\|f\|_p+\|g\|_p$. This is more than a bound on the size of a sum: it lets us measure the distance between two functions by taking the norm of their difference. The resulting distance gives a precise language for convergence and for the geometry of $L^p$.

Throughout, functions are identified when they are equal almost everywhere. This matters because the $L^p$ norm of a function that is zero almost everywhere is zero, even if a chosen representative is nonzero at some points. For $1\leq p<\infty$, the norm is $\|f\|_p=(\int_X |f|^p\,d\mu)^{1/p}$, and $\|f\|_\infty$ is the essential supremum when $p=\infty$. Earlier results establish that $L^p$ is closed under linear combinations and that its norm is definite on these equivalence classes.

Definition (Distance Induced by the $L^p$ Norm): For $f,g\in L^p(X,\mu)$, define $$ d_p(f,g)=\|f-g\|_p. $$ This definition applies for every $1\leq p\leq\infty$.

The difference $f-g$ belongs to $L^p$ by linear closure, so the distance is finite. It is independent of the representatives of $f$ and $g$: changing either on a null set changes their difference only on a null set, and the $L^p$ norm is unchanged under almost-everywhere equality.

Theorem (The $L^p$ Norm Induces a Metric): The function $d_p$ is a metric on $L^p(X,\mu)$, regarded as equivalence classes modulo almost-everywhere equality. In particular, it is nonnegative, symmetric, zero exactly when its arguments are the same equivalence class, and satisfies the triangle inequality $$ d_p(f,h)\leq d_p(f,g)+d_p(g,h). $$

Proof. A norm is nonnegative, so $d_p(f,g)\geq0$. By homogeneity of the norm, $$ d_p(f,g)=\|f-g\|_p=\|-(g-f)\|_p=\|g-f\|_p=d_p(g,f). $$ Definiteness of the $L^p$ norm gives $$ d_p(f,g)=0 \quad\Longleftrightarrow\quad f-g=0\text{ almost everywhere} \quad\Longleftrightarrow\quad f=g\text{ as equivalence classes}. $$ Finally, write $f-h=(f-g)+(g-h)$. Minkowski’s inequality gives $$ d_p(f,h)=\|f-h\|_p \leq\|f-g\|_p+\|g-h\|_p =d_p(f,g)+d_p(g,h). $$ All metric properties follow.

The triangle inequality for the distance is thus exactly Minkowski’s inequality applied to differences. The metric also has a useful invariance: adding the same function to both arguments does not change their distance.

Proposition (Translation Invariance): For all $f,g,u\in L^p(X,\mu)$, $$ d_p(f+u,g+u)=d_p(f,g). $$

Proof. The terms involving $u$ cancel, so $$ d_p(f+u,g+u)=\|(f+u)-(g+u)\|_p=\|f-g\|_p=d_p(f,g). $$ This proves the claim.

Worked Example: Distance on a Two-Point Space

Let $X=\{a,b\}$ have counting measure, and use $p=2$. Define $f(a)=1,\ f(b)=-1$, and $g(a)=2,\ g(b)=1$. Their distance is $$ d_2(f,g)=\left(|1-2|^2+|-1-1|^2\right)^{1/2} =(1+4)^{1/2}=\sqrt{5}. $$ Now let $u(a)=4,\ u(b)=3$. Then $f+u$ has values $5,2$, while $g+u$ has values $6,4$. Directly, $$ d_2(f+u,g+u)=\left(|5-6|^2+|2-4|^2\right)^{1/2} =\sqrt{1+4}=\sqrt{5}. $$ The calculation illustrates translation invariance: the common shift changes both functions but not their separation.

Norm Balls and Convexity

A metric tells us what it means for functions to be near one another. In a normed space, sets of functions within a fixed distance of a center are called balls. The triangle inequality implies that these sets are convex: the line segment between any two points in a ball remains in the ball.

Definition (Norm Ball): For $f\in L^p(X,\mu)$ and $r\geq0$, the closed ball with center $f$ and radius $r$ is $$ \overline{B}_p(f,r)=\{u\in L^p(X,\mu):\|u-f\|_p\leq r\}. $$ For $r>0$, the open ball is $B_p(f,r)=\{u\in L^p(X,\mu):\|u-f\|_p<r\}$.

Theorem (Convexity of $L^p$ Balls): Every closed $L^p$ ball is convex. Every open $L^p$ ball of positive radius is also convex.

Proof. Fix a center $f$. Let $u,v$ belong to the closed ball of radius $r$, and choose $t\in[0,1]$. The convex combination $w=(1-t)u+tv$ satisfies $$ w-f=(1-t)(u-f)+t(v-f). $$ Minkowski’s inequality and homogeneity therefore give $$ \|w-f\|_p \leq(1-t)\|u-f\|_p+t\|v-f\|_p \leq(1-t)r+tr=r. $$ Thus $w$ belongs to the closed ball. If $u,v$ lie in an open ball of radius $r>0$, then $\|u-f\|_p<r$ and $\|v-f\|_p<r$. Consequently, $$ \|w-f\|_p \leq(1-t)\|u-f\|_p+t\|v-f\|_p< (1-t)r+tr=r. $$ The strict inequality holds because both endpoint bounds are strict, including when $t=0$ or $t=1$. Hence $w$ belongs to the open ball as well.

Worked Example: An Average Remains in a Ball

On $[0,1]$ with Lebesgue measure, take $p=2$, center $f=0$, and radius $1$. Let $u(x)=1$ and $v(x)=-1$. Since $$ \|u\|_2=\left(\int_0^1 1\,dx\right)^{1/2}=1, \qquad \|v\|_2=\left(\int_0^1 1\,dx\right)^{1/2}=1, $$ both functions belong to the closed unit ball. Their midpoint is the zero function, and $$ \left\|\frac{u+v}{2}\right\|_2=0\leq1. $$ More generally, the convexity theorem guarantees that every point on the segment from $u$ to $v$ remains in the ball. Here the entire segment consists of constant functions with values between $-1$ and $1$, whose $L^2$ norms are at most $1$.

Convexity is useful when constructing approximations. If two candidate functions satisfy the same norm constraint, then any weighted average of them also satisfies it. The triangle inequality is the key estimate that keeps such averages under control; convexity does not require a separate calculation of the integral defining the norm.

Convergence of Sums and Scalar Multiples

The metric $d_p$ expresses $L^p$ convergence: a sequence $f_n$ converges to $f$ in $L^p$ precisely when $d_p(f_n,f)$ tends to zero. The triangle inequality then controls how convergence behaves under the algebraic operations on functions. In the next result, the scalars are real, as are the functions considered here.

Theorem (Continuity of Addition and Scalar Multiplication): Suppose $f_n\to f$ and $g_n\to g$ in $L^p(X,\mu)$. Then $f_n+g_n\to f+g$ in $L^p$. If also $a_n\to a$ in $\mathbb{R}$, then $a_nf_n\to af$ in $L^p$.

Proof. By Minkowski’s inequality, $$ \|(f_n+g_n)-(f+g)\|_p \leq\|f_n-f\|_p+\|g_n-g\|_p. $$ Both terms on the right tend to zero by the assumed convergence. Therefore the left side tends to zero, proving convergence of the sums.

For scalar multiplication, split the difference as $$ a_nf_n-af=a_n(f_n-f)+(a_n-a)f. $$ Again using Minkowski’s inequality and homogeneity, $$ \|a_nf_n-af\|_p \leq |a_n|\|f_n-f\|_p+|a_n-a|\|f\|_p. $$ A convergent real sequence $(a_n)$ is bounded, so there is a finite constant $C$ such that $|a_n|\leq C$ for all $n$. The first term is then at most $C\|f_n-f\|_p$, which tends to zero. The second term tends to zero because $|a_n-a|\to0$ and $\|f\|_p$ is finite. Thus $\|a_nf_n-af\|_p\to0$, as required.

Worked Example: Convergence of Sums

On $[0,1]$ with Lebesgue measure and $p=2$, set $f_n(x)=x/n$, $g_n(x)=1+x/n$, $f(x)=0$, and $g(x)=1$. Direct integration gives $$ \|f_n-f\|_2 =\left(\int_0^1\frac{x^2}{n^2}\,dx\right)^{1/2} =\frac{1}{\sqrt{3}n}, \qquad \|g_n-g\|_2=\frac{1}{\sqrt{3}n}. $$ Both distances tend to zero. The sums satisfy $f_n+g_n=1+2x/n$ and $f+g=1$, so $$ \|(f_n+g_n)-(f+g)\|_2 =\left(\int_0^1\frac{4x^2}{n^2}\,dx\right)^{1/2} =\frac{2}{\sqrt{3}n}\longrightarrow0. $$ This is the sum-continuity estimate in a concrete case.

Worked Example: A Varying Scalar and Function

On $[0,1]$ with Lebesgue measure, let $f_n(x)=1+x/n$, $f(x)=1$, $a_n=2+1/n$, and $a=2$, using $p=2$. The function sequence converges because $$ \|f_n-f\|_2 =\left(\int_0^1\frac{x^2}{n^2}\,dx\right)^{1/2} =\frac{1}{\sqrt{3}n}. $$ Also $a_n\to2$. The products satisfy $$ a_nf_n-af =(2+1/n)(1+x/n)-2 =\frac{1+2x}{n}+\frac{x}{n^2}. $$ Consequently, $$ \|a_nf_n-af\|_2 \leq \frac{1}{n}\|1+2x\|_2+\frac{1}{n^2}\|x\|_2 \longrightarrow0, $$ where both norms on the right are finite because the functions are bounded on $[0,1]$. Thus the products converge in $L^2$, even though both the scalar and the function vary.

Why These Consequences Matter

The metric viewpoint organizes several facts that otherwise look unrelated. The distance between functions is unchanged by a common translation; norm balls have a stable convex shape; and small changes in inputs lead to small changes in sums and scalar multiples. These are basic tools for working with limits of functions and for studying sets defined by norm bounds.

A useful caution is that convergence in $L^p$ is convergence in the norm, not necessarily pointwise convergence at every point. The estimates here compare norms of differences; by themselves they do not assert pointwise behavior. They do show that algebraic operations preserve $L^p$ limits, which is essential when manipulating sequences in these spaces.

Key takeaway: Minkowski’s inequality makes $d_p(f,g)=\|f-g\|_p$ a genuine distance on almost-everywhere equivalence classes. Its triangle inequality also yields convex norm balls and continuity of addition and scalar multiplication in $L^p$.

Check Your Understanding

Use the metric and continuity results to answer the following questions.

Why must functions in $L^p$ be treated as equivalence classes when $d_p$ is to be a metric?
Which application of Minkowski’s inequality proves the triangle inequality for $d_p$?
In the convexity proof, where is the condition $t\in[0,1]$ used?
Why is boundedness of the convergent scalar sequence $(a_n)$ needed to prove $a_nf_n\to af$?
Does convergence in $L^p$, as used here, mean pointwise convergence at every point? Explain what quantity must instead tend to zero.

What You'll Learn