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Use the periodic modulus of continuitySubdividing a displacement $t$ into at most $\lceil |t|/\delta \rceil$ steps of size at most $\delta$ proves the chaining inequalityFor $|t|\le \delta$, the sharper bound $\omega (f,\delta )$ holds directly.

Set $\delta =n^{-\alpha }$, where $0<\alpha <1$ and $n\ge 1$. Positivity and unit normalized mass of the Fejér kernel giveThe portion $|t|\le \delta$ is at most $\omega (f,\delta )$. For $0<|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$ and $|\sin (nt/2)|\le 1$, soApply this bound and the chaining inequality to the two tails:Let $\beta =1-\alpha >0$. Calculus gives ${\sup }_{u\ge 1}{u}^{-\beta }\log u=1/(e\beta )$, attained at $u=e^{1/\beta }$, and $u^{-\beta }\le 1$. Thus a possible explicit constant for the fractional-scale Fejér approximation bound isThe constant depends only on $\alpha$, not on $f$ or $n$. Since a continuous periodic function is uniformly continuous, the right-hand side tends to zero. The logarithmic tail estimate is what permits every $\alpha <1$; a crude global-oscillation bound would not prove the whole requested range.

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 $[0,1]$.