Scaling And Addition In Lp Norms

Scaling and Addition: Two Different Rules

Scaling a function changes its $L^p$ norm by the absolute value of the scalar. Addition is different: in general, the $p$th power of the norm of a sum is not the sum of the $p$th powers of the individual norms. Minkowski’s inequality gives a bound for addition, while disjoint supports give an exact formula for finite $p$. Keeping these rules separate prevents a common error in $L^p$ calculations.

Throughout, $(X,\mathcal{M},\mu)$ is any measure space; no finiteness, completeness, or sigma-finiteness assumption is needed here. Functions are measurable and real-valued or complex-valued, and functions in $L^p(\mu)$ are identified when they agree $\mu$-a.e. For $1\le p<\infty$, membership means $\int_X |f|^p\,d\mu<\infty$, and $\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}$. For $p=\infty$, membership means that $f$ is essentially bounded, and $\|f\|_\infty$ is its essential supremum in absolute value. All norm equalities below are equalities of real numbers; pointwise function identities hold everywhere for chosen representatives or $\mu$-a.e. as specified.

Scaling a Function

Proposition (Scaling): Let $f\in L^p(\mu)$, where $1\le p<\infty$, and let $c$ be a real or complex scalar. Then $cf\in L^p(\mu)$ and $\|cf\|_p=|c|\|f\|_p$. For $f\in L^\infty(\mu)$, the same conclusion holds: $\|cf\|_\infty=|c|\|f\|_\infty$. These statements include $c=0$.

Proof. For finite $p$, the pointwise identity $|cf|^p=|c|^p|f|^p$ gives $\int_X|cf|^p\,d\mu=|c|^p\int_X|f|^p\,d\mu$. If $c=0$, both sides are zero. If $c\ne0$, taking the $p$th root gives the claimed norm identity. For $p=\infty$ and $c\ne0$, $|cf|\le K$ a.e. holds exactly when $|f|\le K/|c|$ a.e.; taking the least such essential bound gives $\|cf\|_\infty=|c|\|f\|_\infty$. If $c=0$, $cf=0$ everywhere and its essential supremum is zero. $\square$

Worked Example: Scaling a Function in L³

On $[0,2]$ with Lebesgue measure, define $f=2$ on $[0,1]$ and $f=-1$ on $(1,2]$. Compute $\|f\|_3$ and $\|-2f\|_3$.

The two intervals have measure $1$ each, so direct integration gives

\|f\|_3^3=\int_0^2|f|^3\,dm =2^3\cdot 1+|-1|^3\cdot 1=8+1=9.

Thus $\|f\|_3=\sqrt[3]{9}$. The scaling proposition gives $\|-2f\|_3=|-2|\sqrt[3]{9}=2\sqrt[3]{9}$. Directly, the cube of that norm is

\int_0^2|-2f|^3\,dm =4^3\cdot 1+2^3\cdot 1=64+8=72.

Since $(2\sqrt[3]{9})^3=8\cdot9=72$, the direct computation checks the scaling formula.

Addition and Minkowski’s Bound

For $1\le p\le\infty$, Minkowski’s inequality states that if $f,g\in L^p(\mu)$, then $f+g\in L^p(\mu)$ and

\|f+g\|_p\le \|f\|_p+\|g\|_p.

This is the Minkowski inequality established in the earlier material on $L^p$ norms; the statement applies on any measure space, including when either norm is zero. Combining it with the scaling proposition gives the useful weighted form: for scalars $a,b$,

\|af+bg\|_p\le |a|\|f\|_p+|b|\|g\|_p.

Indeed, apply Minkowski to $af$ and $bg$, then apply the scaling identity to each. Repeated application gives the corresponding bound for any finite sum. This is a bound, not usually an equality, and it does not claim that the $p$th powers add.

Worked Example: An Overlapping Sum in L²

On $[0,1]$ with Lebesgue measure, let $f=\chi_{[0,1]}$ and $g=\chi_{[0,1/2]}$. Compute the norm of their sum and compare it with Minkowski’s bound.

First, $\|f\|_2^2=\int_0^1 1\,dm=1$, so $\|f\|_2=1$. Also, $\|g\|_2^2=\int_0^{1/2}1\,dm=1/2$, so $\|g\|_2=1/\sqrt{2}$. On $[0,1/2]$, $f+g=2$; on $(1/2,1]$, $f+g=1$. Therefore

\|f+g\|_2^2 =|2|^2\cdot\frac12+|1|^2\cdot\frac12 =2+\frac12=\frac52, \qquad \|f+g\|_2=\sqrt{\frac52}.

Minkowski’s bound is $1+1/\sqrt{2}$. Both sides are nonnegative, and their squares are $5/2$ and $3/2+\sqrt{2}$, respectively. Since $5/2\le3/2+\sqrt{2}$ is equivalent to $1\le\sqrt{2}$, the computed norm satisfies the bound strictly. The supports overlap on a set of measure $1/2$, so the disjoint-support formula below does not apply.

Disjoint Supports Give Exact Addition

For a measurable function $f$, write $E_f=\{x\in X:f(x)\ne0\}$, its nonzero set for the chosen representative. The functions $f$ and $g$ have disjoint supports a.e. here means $\mu(E_f\cap E_g)=0$. This condition is independent of changes to the functions on null sets.

Proposition (Exact addition for disjoint supports): Let $1\le p<\infty$, and let $f,g\in L^p(\mu)$ have disjoint supports a.e. Then $$ \|f+g\|_p^p=\|f\|_p^p+\|g\|_p^p. $$ More generally, for scalars $a,b$, $$ \|af+bg\|_p^p=|a|^p\|f\|_p^p+|b|^p\|g\|_p^p. $$ For $p=\infty$, the corresponding formula is $\|af+bg\|_\infty=\max\{|a|\|f\|_\infty,\ |b|\|g\|_\infty\}$.

Proof. Outside the null set $E_f\cap E_g$, at most one of $f(x)$ and $g(x)$ is nonzero. Thus, a.e., $|af(x)+bg(x)|^p=|a|^p|f(x)|^p+|b|^p|g(x)|^p$. Integrating this identity and using additivity of the integral for nonnegative integrable functions proves the finite-$p$ formula. The integrals are finite because $f,g\in L^p(\mu)$ and the scalars are finite; taking a $p$th root gives a finite norm.

For $p=\infty$, the absolute value of $af+bg$ equals $|a f|$ on $E_f\setminus E_g$, equals $|b g|$ on $E_g\setminus E_f$, and is zero outside their union, all a.e. Its essential supremum is at most the maximum in the formula. Conversely, $|af|\le|af+bg|$ a.e. on $E_f\setminus E_g$, while $af=0$ on $E_g$ and outside $E_f$; hence its essential supremum cannot exceed that of $af+bg$. The same reasoning applies to $bg$. The essential supremum of the sum is therefore at least each of the two norms and at least their maximum. This proves equality, including when either scalar or function is zero. $\square$

Worked Example: Exact L² Addition on Disjoint Sets

On $[0,2]$ with Lebesgue measure, let $f=3\chi_{[0,1]}$ and $g=4\chi_{(1,2]}$. Compute $\|f+g\|_2$ both from the disjoint-support formula and directly.

The intersection of the two nonzero sets is empty, and each has measure $1$. Thus $\|f\|_2^2=3^2\cdot1=9$ and $\|g\|_2^2=4^2\cdot1=16$. The exact formula gives

\|f+g\|_2^2=9+16=25,\qquad \|f+g\|_2=5.

Directly, $f+g=3$ on $[0,1]$ and $f+g=4$ on $(1,2]$, so $\int_0^2|f+g|^2\,dm=3^2\cdot1+4^2\cdot1=9+16=25$. Taking the square root again gives $5$, as required.

Worked Example: Disjointness in L∞

On $[0,1]$ with Lebesgue measure, set $f=3\chi_{[0,1/2)}$ and $g=4\chi_{[1/2,1]}$. Compute the norm of the sum and compare it with the triangle bound.

The two nonzero sets are disjoint, and both have positive measure $1/2$. Thus $\|f\|_\infty=3$ and $\|g\|_\infty=4$. The disjoint-support maximum formula gives $\|f+g\|_\infty=\max\{3,4\}=4$. Directly, the sum takes values $3$ and $4$ on sets of measure $1/2$, so its essential supremum is $4$. The triangle bound is $3+4=7$, which is valid but not exact.

Choosing the Right Addition Rule

For finite $p$, disjoint supports yield addition of $p$th powers, not addition of the norms themselves. For general supports, use Minkowski’s inequality unless a sharper calculation is available. At $p=\infty$, disjoint supports yield a maximum, not a sum of powers. In every case, scaling is exact and can be applied before an addition estimate.

Common mistake. In the overlapping $L^2$ example, one might incorrectly write $\|f+g\|_2^2=\|f\|_2^2+\|g\|_2^2=1+1/2=3/2$. Direct integration gives $5/2$, because on the overlap the sum has value $2$, whose square is $4$, not $1+1=2$. The exact power-addition rule requires disjoint supports a.e. Without that hypothesis, use Minkowski’s bound or compute the integral of the sum directly.

Check Your Understanding

For each question, specify the relevant value of $p$ and check whether a disjoint-support hypothesis is available.

If $f\in L^4(\mu)$ and $c=-3$, what is $\|cf\|_4$? Give the identity that justifies the answer.
Suppose $f,g\in L^3(\mu)$ have disjoint supports a.e. Write $\|2f-g\|_3^3$ in terms of $\|f\|_3$ and $\|g\|_3$.
For arbitrary $f,g\in L^2(\mu)$, state the general bound for $\|f+g\|_2$. Does it assert equality?
If $f,g\in L^\infty(\mu)$ have disjoint supports a.e., what is $\|f+g\|_\infty$?
Give a reason the $p$th-power addition formula need not hold when the nonzero sets overlap on a set of positive measure.