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The Takagi function is a continuous nowhere differentiable function. The triangular summands are continuous and the geometric tail is bounded uniformly. On dyadic intervals of length $2^{-N}$, all terms of index at least $N$ vanish at both endpoints, while every lower summand has slope $+1$ or $-1$. Each refinement therefore changes the secant slope by exactly one in absolute value. The secants straddling a differentiability point lemma rules out a derivative at every real point.

The bound $0\le d_n\le 2^{-n-1}$ proves uniform convergence of the series by the Weierstrass M-test. Each summand is continuous and satisfies $d_n(x+1)=d_n(x)$, so its sum $t$, the Takagi function, is continuous and has period one.

For $n\ge m$, both $2^n{u}_m$ and $2^n{v}_m$ are integers, giving $d_n(u_m)=d_n(v_m)=0$. ThereforeThe corners of $d_n$ occur on the grid $2^{-n-1}\mathbb{Z}$. When $n<m$, that grid is contained in $2^{-m}\mathbb{Z}$, so the interior of $[u_m,v_m]$ contains no corner of $d_n$. Its slope on this interval is consequently $+1$ or $-1$.

The interval $[u_{m+1},v_{m+1}]$ is a half of $[u_m,v_m]$. All the old slopes persist on this half, and the new summand contributes one more slope in $\{\pm 1\}$. ThusIf $t$ had a finite derivative $L$ at $x_0$, write $t(y)=t(x_0)+L(y-x_0)+r(y)$ with $r(y)=o(|y-x_0|)$. Since $u_m\le x_0<v_m$ and both endpoints tend to $x_0$,Indeed $|v_m-x_0|+|u_m-x_0|=v_m-u_m$; when $u_m=x_0$, its remainder is zero. This is the secants straddling a differentiability point argument, including dyadic $x_0$. But the displayed increments prevent $S_m$ from converging. Hence $\boxed{t\text{ is nowhere differentiable}}$, using periodicity to cover all real $x_0$.

For the distributional derivative, set $r_n(x)=s(2^{n}x)$, using $s$ from part (iii). The change of variable in the distributional derivative pairing gives $d_n^{\prime }=r_n$; these are the Rademacher functions. Uniform convergence implies weak convergence of distributions, and the continuity of distributional differentiation allows termwise differentiation in that topology. Explicitly,This series assertion is a distributional convergence statement. Its remainder satisfies the concrete estimateIt supplies a well-defined distributional derivative even though the ordinary derivative exists nowhere.

Let $\varphi (t)=\mathrm{dist}(t,\mathbb{Z})$. It is continuous and takes values in $[0,1/2]$. The finite sums $f_N(x)={\sum }_{n=1}^N{2}^{-n}\varphi (2^{n}x)$ are continuous and their tails obeyfor every real $x$. To prove continuity without invoking an unproved uniform-limit theorem, fix $x$ and $\epsilon >0$. Choose $N$ with $2\cdot 2^{-N-1}<\epsilon /2$, then use continuity of $f_N$ to make $|f_N(y)-f_N(x)|<\epsilon /2$ near $x$. The triangle inequality then gives $|f(y)-f(x)|<\epsilon$. Thus $f$ is continuous; it is the scaled Takagi function $T(2x)/2$.

For the general secant claim, differentiability gives $g(x+h)=g(x)+g^{\prime }(x)h+r(h)$, where $|r(h)|\le \epsilon |h|$ for sufficiently small $h$, with $r(0)=0$. Since the two endpoints straddle $x$,Hence the secant limit is $g^{\prime }(x)$, including cases where an endpoint equals $x$. This proves secants straddling a differentiability point.

Now take the nested dyadic intervals $u_N=2^{-N}\lfloor 2^{N}x\rfloor$, $v_N=u_N+2^{-N}$. They straddle $x$ and shrink to it. For every $n\ge N$, $2^n{u}_N$ and $2^n{v}_N$ are integers, so the corresponding summand vanishes at both endpoints. For $n<N$, all corners of that summand lie on the dyadic grid of mesh $2^{-N}$; on the chosen open interval it is linear with slope ${\epsilon }_n\in \{-1,1\}$. Therefore its total secant slope isRefining to the nested interval preserves the slopes of all old terms and adds exactly one new term of slope $+1$ or $-1$. Thus $|S_{N+1}-S_N|=1$ for every $N$, so $S_N$ cannot converge. This contradicts the secant consequence of differentiability at any $x$. Therefore $f$ is nowhere differentiable. At dyadic $x$ the intervals chosen on its right satisfy the same argument. The printed reference to a denominator $2^{-n}$ is naturally read as mesh $2^{-n}$, or denominator $2^n$, as used here.

Use the local linear approximation from part (a), with $L=f^{\prime }(c)$. Since $x_n\le c\le y_n$ and $x_n\ne y_n$, we have $y_n-x_n>0$, and subtraction givesThe coefficients $(y_n-c)/(y_n-x_n)$ and $(c-x_n)/(y_n-x_n)$ are nonnegative and sum to one. ThereforeHere $ϵ(c)=0$ handles any endpoint of a secant that happens to equal $c$. Thus secants straddling a differentiability point satisfy