test

设函数 \(f\) 在有界闭区间 \([a,b]\) 上连续，在 \((a,b)\) 上可微，则存在 \(\xi\in (a,b)\) 使得

\[f'(\xi)=\frac{f(b)-f(a)}{b-a}. \]

设函数 \(f\) 在有界闭区间 \([a,b]\) 上连续，在 \((a,b)\) 上可微，则存在 \(\xi\in (a,b)\) 使得

\[f'(\xi)=\frac{f(b)-f(a)}{b-a}. \]

任何大于2的偶数可以表示为两个质数之和（哥德巴赫猜想）。

若一个函数在闭区间上连续，则它在该区间上有界。

有限维空间中的线性算子必有界。

此结论在无限维空间中不成立。

显然。

读者自证。

First, let's truncating $T$ at depth $\log\left(\frac{s}{\epsilon}\right)$. Each leaf at depth more than $\log\left(\frac{s}{\epsilon}\right)$ provides error with probability at most $2^{-\log(s/\epsilon)}=\frac{\epsilon}{s}$. So the total error is at most $\epsilon$. We assume that $T$ has depth at most $\log\left(\frac{s}{\epsilon}\right)$ in the following.

Clearly, a decision tree with \(s\) leaves can be represented by union of \(s\) "AND" terms, denoted by \(f\). Since every "AND" term has \(L_1\le 1\) and at most \(\log\left(\frac{s}{\epsilon}\right)\) variables, so \(L_1(f)\le s\) and the degree of \(f\) is at most \(\log\left(\frac{s}{\epsilon}\right)\).

Finally, let \(h\) be the truncation of \(f\) by keeping only the Fourier coefficients with \(|\widehat{f}_S|\ge\frac{\epsilon}{L_1(f)}\). Then \(L_0(h)\le \frac{L_1(f)}{\frac{\epsilon}{L_1(f)}}\le\frac{s^2}{\epsilon}\). By Parseval's identity, the missing terms have contribution at most

\[\sum_{|\widehat{f}_S|<\frac{\epsilon}{L_1(f)}} \left(\widehat{f}_S\right)^2 \le \max_{|\widehat{f}_S|<\frac{\epsilon}{L_1(f)}} |\widehat{f}_S|\cdot\sum_{|\widehat{f}_S|<\frac{\epsilon}{L_1(f)}} |\widehat{f}_S|\le L_1(f)\cdot \frac{\epsilon}{L_1(f)}=\epsilon. \]

Thus, \(h\) are \(\epsilon\)-close to \(f\). By triangle inequality, we have

\[\mathbb{E}_{\boldsymbol{x}\sim \mathcal{D}}[T(\boldsymbol{x})- h(\boldsymbol{x})]^2\le 2\mathbb{E}_{\boldsymbol{x}\sim \mathcal{D}}\left[(T(\boldsymbol{x})- f(\boldsymbol{x}))^2+(f(\boldsymbol{x})- h(\boldsymbol{x}))^2\right]\le 4\epsilon. \]