= Solution
Put $L=2\pi$ and represent the <circle> by $[0,L)$. The full <orthonormal basis> of <Haar wavelets> includes the constant $H_*=L^{-1/2}$ and, for every dyadic interval $I$ of length $\ell=L2^{-j}$, the normalized detail
$$
H_I=\ell^{-1/2}(\mathbf1_{I_{\mathrm{left}}}-\mathbf1_{I_{\mathrm{right}}}),\qquad c_I=\int_0^Lf(t)\overline{H_I(t)}\,dt.
$$
Take half-open intervals and the corresponding values at their boundaries, so every point belongs to exactly one interval at each level. The constant basis function is necessary: if it were omitted, even $f\equiv1$ could not satisfy the asserted convergence. We use the full conventional <Haar wavelet> system in the threshold sum.
Let $A_Jf$ be the function equal to the mean of $f$ on each level-$J$ dyadic interval. Comparing the means on a parent and its two children proves the <Haar refinement identity> directly: the difference between the child mean and the parent mean is the appropriate value of $c_IH_I$. Inductively this gives the <Haar projection>
$$
A_Jf=c_*H_*+\sum_{j=0}^{J-1}\sum_{I\text{ at level }j}c_IH_I,\qquad c_*H_*=\frac1L\int_0^Lf(t)\,dt.
$$
Writing $\omega_f(s)=\sup_{\operatorname{dist}(x,y)\leq s}|f(x)-f(y)|$ for the <modulus of continuity>, the cell-average formula yields
$$
\|A_Jf-f\|_\infty\leq\omega_f(L2^{-J})\longrightarrow0.
$$
For a detail on a cell $I$, its integral is zero. Subtracting $f(x_I)$ for any $x_I\in I$ therefore gives the coefficient estimate
$$
|c_I|\leq\int_I|f(t)-f(x_I)||H_I(t)|\,dt\leq\omega_f(\ell)\sqrt\ell.
$$
In particular $|c_I|\leq2\|f\|_\infty\sqrt\ell$, so for every $\delta>0$ only finitely many details can have $|c_I|\geq\delta$. The hard-threshold sum
$$
T_\delta f=\sum_{|c_H|\geq\delta}c_HH
$$
is thus a well-defined finite sum. Its chosen details need not form a complete set of levels; this is why convergence of the <Haar projections> alone does not prove the result.
Fix $\eta>0$. Choose $J\geq1$ so that $\omega_f(L2^{-J})\leq\eta$. All finer details satisfy $|c_I|\leq\eta\sqrt L\,2^{-j/2}$ at level $j\geq J$. For sufficiently small $\delta$, every nonzero coefficient at levels below $J$, and the constant coefficient if nonzero, is retained. Also choose the unique integer $K\geq J$ such that
$$
\frac{\delta2^{K/2}}{\sqrt L}\leq\eta<\frac{\delta2^{(K+1)/2}}{\sqrt L}.
$$
For $j>K$ we have $|c_I|\leq\eta\sqrt L2^{-j/2}<\delta$, so no such detail is retained. Consequently $T_\delta f$ differs from $A_{K+1}f$ only by omitted details at levels $J$ through $K$. At any point there is at most one supported detail per level, and an omitted detail has magnitude at most $\delta/\sqrt{L2^{-j}}$. The <geometric bound for omitted Haar details> gives
$$
\|T_\delta f-A_{K+1}f\|_\infty\leq\sum_{j=J}^K\frac{\delta2^{j/2}}{\sqrt L}\leq\frac{\eta}{1-2^{-1/2}}.
$$
Since $\|A_{K+1}f-f\|_\infty\leq\eta$, we conclude
$$
\|T_\delta f-f\|_\infty\leq\left(1+\frac1{1-2^{-1/2}}\right)\eta.
$$
The right side can be made arbitrarily small, proving <uniform hard-threshold convergence of normalized Haar expansions>:
$$
\boxed{\left\|\sum_{|\widehat f(H)|\geq\delta}\widehat f(H)H-f\right\|_\infty\longrightarrow0\quad(\delta\downarrow0).}
$$
The square-root scaling imposed by the specified $L^2$ normalization is what makes the omitted contributions a controlled <geometric series>.
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