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For a function $f:[0,1]\to \mathbb{R}$ and an integer $n\ge 1$, its Bernstein polynomial isThe Bernstein basis supplies nonnegative weights whose sum is one. Equivalently, the weights are the probabilities of a binomial distribution, and the result is an expected value. At a given $x$, this averages the sampled values $f(k/n)$ with weights concentrated near $k/n=x$.^[ref]

The endpoints are exact: $B_{n}f(0)=f(0)$ and $B_{n}f(1)=f(1)$. Interior sampled points need not be interpolated. For example, binomial distribution moments give $B_n(x)=x$ and $B_n(x^2)=x^2+x(1-x)/n$. Thus every affine function is reproduced exactly, whereas $x^2$ is generally changed. Bernstein polynomial degree preservation is a statement about degree of a polynomial, not equality of the entire polynomial.

For a continuous $f$, the uniform continuity of $f$ and the variance $\mathrm{Var}(K/n)=x(1-x)/n\le 1/(4n)$ explain uniform convergence. Split the expected value of the error into $|K/n-x|<\delta$, where uniform continuity makes the change small, and its complement, whose probability is at most $1/(4n{\delta }^2)$ by Chebyshev's inequality. The bounded function $f$ bounds the remaining error. This proves convergence without assuming differentiability.

For $f\in C([0,1{]}^d)$, the tensor-product Bernstein polynomial is the expected value of $f(X_1/n,\dots ,X_d/n)$ for independent random variables $X_j\sim \mathrm{Bin}(n,x_j)$. The same uniform continuity and coordinatewise variance estimates give uniform convergence for each fixed finite $d$.

For $n\ge 1$, the Bernstein polynomial isThe weights sum to one by the binomial theorem. Equivalently, if $K$ has the binomial distribution with parameters $n,x$, then $B_n(f;x)=\mathbb{E}f(K/n)$. This interpretation makes the cancellation of apparent high-degree terms transparent.

For the falling factorial $u^{\underline{r}}=u(u-1)\cdots (u-r+1)$, with $u^{\underline{0}}=1$, direct cancellation in the binomial weights givesThe polynomials $u^{\underline{0}},\dots ,u^{\underline{m}}$ form a monic triangular basis of the degree-at-most-$m$ polynomials. Hence $u^m=u^{\underline{m}}+{\sum }_{r<m}{c}_r{u}^{\underline{r}}$, andFor $m\le n$, its leading coefficient is $n^{\underline{m}}/n^m\ne 0$. If $p$ has actual degree $m$ and leading coefficient $a_m\ne 0$, linearity shows that the leading coefficient of $B_{n}p$ is $a_m{n}^{\underline{m}}/n^m$; lower-degree terms cannot cancel it. Thus the Bernstein polynomial degree preservation property givesThis includes the requested $m<n$, and also explains why the property cannot simply be extended to $m>n$.

The same moments give $\mathbb{E}K=nx$ and $\mathbb{E}{K}^2=n(n-1)x^2+nx$. ThereforeThe first two approximation errors vanish identically, whileAll three convergences are consequently uniform convergence on $[0,1]$.