Why the Largest Error Costs More
Integrated squared error averages performance over the covariate space. Supremum-norm error instead asks how badly an estimator performs at its worst location. Even if errors in individual regions are modest, taking the largest across many regions makes large fluctuations more likely. In regular Hölder regression settings, this distinction introduces a logarithmic factor into the expected squared supremum-norm rate.
We work with a regression function \(m\) on \([0,1]^d\), with smoothness \(0<\beta<1\). The results below use a regular fixed design and sub-Gaussian noise for the upper bound, and Gaussian noise for the lower bound. This is a deliberately specified setting in which both the rate and the source of the logarithm can be seen directly.
For definiteness, let \(\mathcal H^\beta(L)\) be the Hölder class of functions satisfying \(\|m\|_\infty\le L\) and \(|m(x)-m(y)|\le L\|x-y\|_2^\beta\) for all \(x,y\in[0,1]^d\). In the regression model, \(Y_i=m(x_i)+\varepsilon_i\), where the design points \(x_i\) are fixed. Assume the errors are independent, have mean zero, and are \(\sigma\)-sub-Gaussian: \(E[\exp(t\varepsilon_i)]\le \exp(\sigma^2t^2/2)\) for every real \(t\). The design is regular in the sense that partitioning the cube into equal cubes of side \(h\) gives between \(cnh^d\) and \(Cnh^d\) design points in every cube, for fixed positive constants \(c,C\), at the resolutions used below.
The Rate and Its Comparison
Under these conditions, the characteristic expected squared supremum-norm rate is \((\log n/n)^{2\beta/(2\beta+d)}\). The corresponding integrated squared-error rate under regular design is \(n^{-2\beta/(2\beta+d)}\). Here integrated risk means \(E_m[\int_{[0,1]^d}(\hat m(x)-m(x))^2\,dx]\). The powers of \(n\) agree, but supremum loss has an extra logarithmic factor.
The upper bound balances two contributions. A partition with side length \(h\) approximates a Hölder function to order \(h^\beta\), so squared bias is of order \(h^{2\beta}\). There are about \(h^{-d}\) cells, and the largest noise fluctuation among their averages has squared size of order \(\log(1/h)/(nh^d)\). Balancing these terms yields the logarithmic rate.
Worked Example: Balancing Bias and Maximum Noise
Take \(d=2\) and \(\beta=1/2\). For cell side length \(h\), the squared approximation error is of order \(h^{2\beta}=h\). Each cell contains order \(nh^2\) observations, and the maximum squared noise level is of order \(\log(1/h)/(nh^2)\). Equating the two orders gives
Consequently \(h\) is of order \((\log n/n)^{1/3}\), and the squared supremum-norm risk is of order \(h=(\log n/n)^{1/3}\). The integrated squared-error rate in this case is \(n^{-1/3}\), without the logarithmic multiplier.
Worked Example: Comparing Rates at a Given Sample Size
Let \(d=1\), \(\beta=1/2\), and \(n=1000\). The exponent \(2\beta/(2\beta+d)\) is \(1/(1+1)=1/2\). Ignoring constants, the integrated squared-error benchmark is
The supremum-norm benchmark uses \(\log(1000)\approx 6.908\):
These are rate calculations, not predictions of exact risk: the omitted constants can be substantial. The comparison illustrates that the logarithmic factor changes the finite-sample benchmark, even though both rates have the same power \(n^{-1/2}\).
Upper Bound: Controlling the Maximum Over Cells
Divide \([0,1]^d\) into \(K\) equal cubes of side \(h\), where \(K=h^{-d}\). On each cube, define \(\hat m\) to be the average of the responses whose design points fall in that cube. If \(x\) lies in a cube \(B\), the estimator’s mean on that cube is the average of \(m(x_i)\) over its design points. Every such \(x_i\) is within distance \(\sqrt d\,h\) of \(x\), so the Hölder condition gives
Write the centered noise average in cell \(B\) as \(Z_B\), and let \(N_B\) be its number of design points. Independence and the sub-Gaussian assumption imply \(P(|Z_B|>t)\le 2\exp(-N_Bt^2/(2\sigma^2))\). Since there are \(K\) cells and \(N_B\ge cnh^d\), the union bound gives
Integrating this tail bound yields \(E[\max_B|Z_B|^2]\le C_1\sigma^2\log(2K)/(nh^d)\) for a constant \(C_1\). For example, split the integral \(E[W^2]=\int_0^\infty P(W^2>s)\,ds\), with \(W=\max_B|Z_B|\), at \(s_0=2\sigma^2\log(2K)/(cnh^d)\). The integral below \(s_0\) is at most \(s_0\); above \(s_0\), the displayed exponential tail integrates to at most \(2\sigma^2/(cnh^d)\). Thus the stated bound follows.
For any \(x\), the estimation error is the sum of its cell’s bias and noise average. The inequality \((a+b)^2\le 2a^2+2b^2\) therefore gives
Choose \(h\) of order \((\log n/n)^{1/(2\beta+d)}\), rounding to an available partition resolution. Then \(h^{2\beta}\) has order \((\log n/n)^{2\beta/(2\beta+d)}\). Also \(\log(2h^{-d})\) has order \(\log n\), so the variance term has that same order. This proves the upper bound. \(\square\)
Worked Example: Checking the Resolution in One Dimension
Set \(d=1\) and \(\beta=1/2\). The general balancing equation is \(h^{2\beta+d}\asymp\log n/n\). Here \(2\beta+d=2\), so
Squaring the bias order \(h^\beta\) gives \(h^{2\beta}=h\), hence a risk of order \((\log n/n)^{1/2}\). The largest noise average has squared order \(\log(1/h)/(nh)\). Since \(h\) has the displayed order and \(\log(1/h)\) is of order \(\log n\), this is also of order \((\log n/n)^{1/2}\). Both terms agree, as required by the balancing choice.
Lower Bound: Many Possible Locations for a Difficult Feature
The logarithm is not merely a weakness of the histogram estimator. To see why it is necessary, construct many regression functions that each contain one small, localized bump, with the bump placed in a different region. Each function is separated from the zero function by the same amount in supremum norm. Yet when the bump is narrow, its effect on the data is limited to a small part of the design. Distinguishing among many possible bump locations requires enough information to identify which location is active.
Here are the details of the lower-bound argument. Choose a fixed smooth function \(\phi\), supported strictly inside the unit cube, with \(\|\phi\|_\infty=1\). At a small scale \(h\), place \(M\) disjoint cubes inside \([0,1]^d\), with \(M\) of order \(h^{-d}\), and let \(z_j\) be the lower corner of cube \(j\). For a sufficiently small constant \(a>0\), define
The supports are disjoint, and scaling the bump by \(h^\beta\) keeps its Hölder-\(\beta\) constant bounded uniformly in \(h\). Choosing \(a\) small enough places every \(m_j\) in \(\mathcal H^\beta(L)\). Their pairwise supremum distances are \(a h^\beta\). If the errors are Gaussian with variance \(\sigma^2\), the Kullback–Leibler divergence from the data distribution under \(m_j\) to that under the zero function is
The inequality follows because the bump is bounded by \(a h^\beta\) and its support contains at most a constant times \(nh^d\) design points. Take \(h\) of order \((\log n/n)^{1/(2\beta+d)}\). Then \(\log M\) is of order \(\log n\), and the divergence bound is at most a sufficiently small constant times \(\log M\), by choosing \(a\) small enough.
Fano’s inequality states that, for \(M\) possible labels with uniform prior, the probability of incorrectly identifying the label is at least \(1-(I+\log 2)/\log M\), where \(I\) is the mutual information between the label and the data. In this construction, \(I\) is at most the average divergence to the zero-function distribution, so the choice just made ensures a classification error probability bounded away from zero for large \(n\). Given any estimator, classify it by the closest \(m_j\) in supremum norm. If its supremum error is less than \(a h^\beta/2\), the classification is correct, because the alternatives are separated by \(a h^\beta\). Thus its worst-case expected squared supremum error is at least a positive constant times \(h^{2\beta}\). Substituting the chosen \(h\) gives the lower rate \((\log n/n)^{2\beta/(2\beta+d)}\). \(\square\)
What the Logarithm Means in Practice
A useful way to remember the contrast is that integrated loss aggregates errors across locations, whereas supremum loss is sensitive to the worst one. A resolution with \(K\) local regions requires controlling all \(K\) noise averages at once; their maximum typically brings in a factor of order \(\log K\) in squared error. At the optimal resolution, \(K\) grows with \(n\), and \(\log K\) is of order \(\log n\). This is the source of the extra factor in the rate.
The theorem’s assumptions matter. The design must supply enough observations in every region at the selected resolution; a region with very few observations can have much larger noise. The noise tail condition supports the maximum-fluctuation bound, while the lower bound uses Gaussian noise to exhibit a concrete hard subproblem. The stated result is for \(0<\beta<1\); higher smoothness requires approximation methods that use the additional smoothness, while the same logarithmic comparison is the standard one in regular settings.
- Mistake: Reusing the integrated squared-error rate as the supremum-norm rate because their powers of \(n\) match. Correction: The supremum rate also contains \((\log n)^{2\beta/(2\beta+d)}\).
- Mistake: Treating the logarithm as an arbitrary penalty caused only by a crude union bound. Correction: The localized-bump lower bound shows that identifying the worst region creates a genuine minimax cost.
- Mistake: Comparing rates while changing the loss or omitting the square. Correction: Here both displayed rates concern squared error; the supremum loss squares \(\|\hat m-m\|_\infty\).
Check Your Understanding
Use the definitions and rate arguments above to answer these questions.
- For \(d=2\) and \(\beta=1/2\), what is the exponent \(2\beta/(2\beta+d)\), and how do the two squared-error rates differ?
- Why does a maximum over \(K\) cellwise noise averages have a logarithmic contribution even when each individual average has variance of order \(1/(nh^d)\)?
- In the upper-bound proof, what are the bias and maximum-noise contributions before choosing \(h\)?
- Why are the localized alternatives in the lower-bound argument separated in supremum norm even though each bump affects only a small region?
- What design or noise feature could make the stated supremum-norm upper bound fail?