Norm Convergence Is a Different Kind of Convergence
A sequence of functions can converge at every point and still fail to approach its limit in \(L^p\) norm. Conversely, norm convergence need not mean pointwise convergence of the chosen representatives at every point. To make these distinctions precise, it matters that \(L^p\) consists of equivalence classes of functions equal almost everywhere, while pointwise convergence is a statement about particular representatives.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space, with no assumption that \(\mu\) is finite, sigma-finite, or complete. Functions are real-valued or complex-valued and measurable. For \(1\le p<\infty\), \(L^p(\mu)\) is the space of equivalence classes modulo equality \(\mu\)-a.e. of functions \(f\) satisfying \(\int_X |f|^p\,d\mu<\infty\), with norm \(\|f\|_p=(\int_X |f|^p\,d\mu)^{1/p}\). The space \(L^\infty(\mu)\) consists of equivalence classes of essentially bounded measurable functions, with \(\|f\|_\infty=\inf\{C\ge0: |f|\le C\ \mu\text{-a.e.}\}\). The norm is well-defined on equivalence classes.
The Cauchy condition compares every sufficiently late pair of terms; convergence compares each sufficiently late term with one specified limit. The triangle inequality for the \(L^p\) norm, given by Minkowski’s inequality (including its endpoint version for \(p=\infty\)), immediately relates the two notions.
Proof. Suppose first that \(f_n\to f\) in \(L^p(\mu)\). For \(n,m\) sufficiently large, both \(\|f_n-f\|_p<\varepsilon/2\) and \(\|f_m-f\|_p<\varepsilon/2\). The triangle inequality gives \(\|f_n-f_m\|_p\le\|f_n-f\|_p+\|f-f_m\|_p<\varepsilon\), proving the Cauchy condition.
Now let \(p<\infty\). For any measurable \(h\) and any \(\varepsilon>0\), the pointwise inequality \(|h|^p\ge\varepsilon^p\) on \(\{|h|>\varepsilon\}\) gives \(\varepsilon^p\mu(\{|h|>\varepsilon\})\le\int_X|h|^p\,d\mu\). Apply this with \(h=f_n-f\). The right side tends to zero, so the measure of the strict tail \(\{|f_n-f|>\varepsilon\}\) tends to zero. For \(p=\infty\), once \(\|f_n-f\|_\infty<\varepsilon\), the definition of essential supremum implies \(|f_n-f|\le\varepsilon\) almost everywhere; hence \(\mu(\{|f_n-f|>\varepsilon\})=0\). This proves the proposition. \(\square\)
Distances Between Indicators and Step Functions
For indicators, the norm distance is determined exactly by the region on which the sets differ. The formula applies on any measure space, including when the symmetric difference has infinite measure.
Proof. At each point, \(|\chi_A-\chi_B|\) is \(1\) on \(A\,\triangle\,B\) and \(0\) elsewhere. Thus \(|\chi_A-\chi_B|^p=\chi_{A\,\triangle\,B}\), and integrating gives the finite-\(p\) formula. For \(p=\infty\), if the symmetric difference is null, the difference vanishes a.e. and its essential supremum is \(0\). If the symmetric difference has positive measure, the difference has magnitude \(1\) on a positive-measure set and magnitude at most \(1\) everywhere, so its essential supremum is \(1\). \(\square\)
Worked Example: Two Interval Indicators
On \(\mathbb{R}\) with Lebesgue measure \(m\), take \(A=[0,1)\) and \(B=[0,3/2)\). Their symmetric difference is \([1,3/2)\), of measure \(3/2-1=1/2\). Therefore, for finite \(p\), \(\|\chi_A-\chi_B\|_p=(1/2)^{1/p}\); for \(p=\infty\), the norm is \(1\), since the indicators differ on a set of positive measure.
Check: For \(p=1\), integrating the difference’s absolute value gives \(1\cdot(1/2)=1/2\). For \(p=2\), its squared integral is \(1^2\cdot(1/2)=1/2\), so the norm is \(\sqrt{1/2}\). For \(p=\infty\), values of magnitude \(1\) occur on an interval of length \(1/2>0\), confirming the essential supremum is \(1\).
For step functions, divide the domain into disjoint measurable pieces on which the difference is constant, then integrate the \(p\)-th power of its absolute value. At \(p=\infty\), take the largest absolute value occurring on a set of positive measure; values only on null sets do not affect the norm.
Worked Example: A Difference of Step Functions
On \(\mathbb{R}\) with Lebesgue measure, let \(f=2\chi_{[0,1)}+\chi_{[1,2)}\) and \(g=\chi_{[0,1)}+3\chi_{[1,3/2)}\). On \([0,1)\), \(f-g=2-1=1\); on \([1,3/2)\), \(f-g=1-3=-2\); on \([3/2,2)\), \(f-g=1-0=1\); and outside \([0,2)\), the difference is zero. The interval lengths are respectively \(1\), \(1/2\), and \(1/2\).
Thus \(\|f-g\|_1=1+2(1/2)+1(1/2)=5/2\), and \(\|f-g\|_2=\sqrt{1+4(1/2)+1(1/2)}=\sqrt{7/2}\). For \(p=\infty\), the largest magnitude on a positive-length interval is \(2\), so \(\|f-g\|_\infty=2\). Check: the \(p=1\) value is also the sum of the three absolute contributions \(1+1+1/2=5/2\), and the \(p=2\) squared value is \(1+2+1/2=7/2\).
A Moving Endpoint: Cauchy Depends on the Norm
Worked Example: Indicator Functions with Moving Endpoints
On \(\mathbb{R}\) with Lebesgue measure, define \(f_n=\chi_{[0,1+1/n]}\) for \(n\ge1\), and let \(f=\chi_{[0,1]}\). For \(m,n\ge1\), the intervals are nested, and their symmetric difference has length \(|1/n-1/m|\). Hence, for finite \(p\),
Given \(\varepsilon>0\), choose \(N\) with \(N^{-1/p}<\varepsilon\). If \(m,n\ge N\), the displayed bound is at most \(N^{-1/p}<\varepsilon\), so \((f_n)\) is Cauchy in every finite-\(p\) norm. Also \(f_n-f\) is an indicator of an interval of length \(1/n\), up to endpoints, and therefore \(\|f_n-f\|_p=(1/n)^{1/p}\to0\).
In \(L^\infty(\mathbb{R})\), if \(m\ne n\), the symmetric difference has positive length, so \(\|f_n-f_m\|_\infty=1\). Taking \(\varepsilon=1/2\), no \(N\) can make all pairs \(m,n\ge N\) have distance below \(\varepsilon\); the sequence is not Cauchy in \(L^\infty\). Likewise, \(\|f_n-f\|_\infty=1\) for every \(n\), so it does not converge to \(f\) in that norm.
Check: For \(n=2\), the excess interval has length \(1/2\), giving \(\|f_2-f\|_1=1/2\) and \(\|f_2-f\|_2=\sqrt{1/2}\), while the \(L^\infty\) distance is \(1\). These agree with the formulas for each norm.
Pointwise Convergence Does Not Control Norm Distance
Worked Example: Shrinking Spikes
On \([0,1]\) with Lebesgue measure, fix \(1\le p<\infty\) and set \(u_n=n^{1/p}\chi_{(0,1/n)}\). For every \(x\in[0,1]\), \(u_n(x)\to0\): at \(x=0\) every value is zero, and for each \(x>0\), \(x<1/n\) can hold for only finitely many \(n\). Nevertheless,
Thus pointwise convergence to zero does not imply norm convergence to zero. For \(p=\infty\), instead set \(u_n=\chi_{(0,1/n)}\). These functions also converge pointwise to zero, while \(\|u_n\|_\infty=1\) for every \(n\), because each interval \((0,1/n)\) has positive measure. Check: the strict interval \((0,1/n)\) excludes \(x=0\), and its measure is \(1/n>0\); the finite-\(p\) integral is exactly \(n(1/n)=1\).
Even norm convergence need not give pointwise convergence of the full sequence. For example, on \([0,1]\), list in order all \(k\) half-open intervals \([j/k,(j+1)/k)\), \(j=0,\ldots,k-1\), for each block \(k=1,2,\ldots\), and let each function in that block be the indicator of the corresponding interval. For finite \(p\), each norm is \(k^{-1/p}\), which tends to zero as the block number grows. But every \(x\in[0,1)\) belongs to one interval in every block, so its function values are \(1\) infinitely often and \(0\) infinitely often; there is no pointwise limit there. For \(p=\infty\), this sequence does not converge in norm to zero, since every term has norm \(1\).
Finally, pointwise values depend on representatives, not just on an \(L^p\) element. For instance, on \([0,1]\), the functions \(v_n=\chi_{\mathbb{Q}\cap[0,1]}\) for even \(n\), and \(v_n=0\) for odd \(n\), all represent the zero element of every \(L^p\): the rationals are countable and hence have Lebesgue measure zero. Thus \(\|v_n\|_p=0\) for every \(n\), but the chosen values oscillate at every rational point. The \(L^p\) statement is norm convergence of equivalence classes; pointwise convergence requires additional choices of representatives.
A common mistake is to replace “small \(L^p\) norm” with “small value at every point.” For the finite-\(p\) spike above, \(\|u_n\|_p=1\), not a small norm, despite pointwise convergence to zero. More generally, norm estimates control an integral or an essential supremum, not the value at each point. Conversely, the moving-endpoint example is Cauchy in finite \(p\) but not in \(L^\infty\): the same positive-measure discrepancy has a small \(L^p\) norm when its length is small, but its essential supremum remains \(1\).
Check Your Understanding
Use the definitions and compute each norm from the measure of the relevant sets.
- On \(\mathbb{R}\) with Lebesgue measure, compute \(\|\chi_{[0,2)}-\chi_{[1,3)}\|_p\) for finite \(p\) and for \(p=\infty\).
- For \(f_n=\chi_{[0,1+1/n]}\) on \(\mathbb{R}\), show directly that the sequence is Cauchy in \(L^1\) and not Cauchy in \(L^\infty\).
- On \([0,1]\), define \(w_n=n^{1/2}\chi_{(0,1/n)}\). Verify its pointwise limit and compute \(\|w_n\|_2\).
- Explain why the essential supremum of an indicator difference is zero when the indicators differ only on a null set.
- Give a reason that norm convergence is a statement about equivalence classes, whereas pointwise convergence refers to chosen representatives.