15 added 6 characters in body

Your guess is indeed right. Following a similar idea gives you the Takagi or blancmange function. It is even quasi-Lipschitz (it has a modulus of continuity $\omega(t)=ct(|\log(t)|+1)$ for a suitable constant $c>0$), thus it's Hoelder of any positive exponent less than 1. It is not even BV in any open interval, thus $L^p_{loc}$ W^{1,p} _ {loc}$for no$p\geq1$. Rmk 1. The above example is for dimension 1: but of course it holds in any dimension a fortiori. Rmk 2. To get an example with a more classical flavor, actually a Weierstrass function, replace$s(x)$with$\cos(x)$. I'd say that the resulting Fourier series defines a function with the same features, by the same reasons (the function$\cos(x)$works better than$\sin(x),$in view of point 2 below.) Rmk 3. Once you know that the Weierstrass function$f(x):=\sum_{k=0}^\infty 2^{-k}\cos(2^k x)$is nowhere differentiable, you also have that it is BV on no open interval, for BV on an interval would imply differentiability a.e. there. However, for your needs it seems more direct just showing it has infinite variation on any interval. Details. 1. To prove that the Takagi function$f(x)$admits the above modulus of continuity, recall that that$f$is characterized as the fixed point of the affine contraction$T:C_b(\mathbb{R})\to C_b(\mathbb{R})$such that$(Tf)(x)=\frac{1}{2}f(2x)+s(x),$for all$x\in\mathbb{R}$, where$s(x)$is the distance function from the integers (a zig-zag piecewise 1-periodic function). Just find a$c$such that the subset of$C_b(\mathbb{R})$of functions that admit$\omega$as modulus of continuity is a$T$-invariant set. The latter subset is obviously closed and non-empty, so the fixed point is there. (The above illustrated a standard general technique to prove properties of objects found by means of the contraction principle). 2. Proving that$f$is not of bounded variation on$[0,1]$(hence in no open interval, due to the self-similarity encoded in the fixed point equation), requires a small computation on the partial sum$f_n$of the series defining$f$. Let $$f_n(x):=\sum_{k=0}^{n-1}\ 2^{-k} s(2^k x).$$ First note that the derivative of$f_n$only takes integer values, which of course come as a result of the sum of$n$terms$\pm 1$(with all the$2^n$possible signs). In particular, for any$n\in\mathbb{N}$the function$f_{2n}$has${2n \choose n} $flat intervals of lenght$2^{-2n}$within the unit interval$I$, and has derivative larger than$2$in absolute value elsewhere in$I$. Thus, for the subsequent odd index$2n+1,$the function$f_{2n+1}$has${2n \choose n}$local maxima in$I$(located in the mid-points of the above intervals). Moreover, passing to$f_{2n+1}$each maximum point contributes to the increment of the total variation with$2^{-2n}$, while the total variation remains unchanged passing from$2n+1$to the next even index$2n+2$. The conclusion is that, for any$n$, the total variation of$f_n$on$I$is $$V(f_n;I)=\sum_{0\leq k < n/2}{2k\choose k}2^{-2k} =O\big(\sqrt{n}\big),$$ since by the classical asymptotics for the central binomial coefficient,${2k \choose k}=\frac{4^k}{\sqrt{\pi k}}(1+o(1)),\ k\to\infty.$So actually$V(f_n;I)$diverges. Yet this would not be sufficient to conclude that$V(f,I)=\infty,$as the total variation is only lower semicontinuous with respect to the uniform convergence. However, the discrete variation on a given subdivision$P:=\{t_0 < \dots < t_r \}$$$V(f_n; P\ )=\sum_{i=0}^{r-1}\ \big|f_n(t_{i+1})-f(t_i)\big|$$ does of course pass to the limit under even pointwise convergence. Now the point is that, for the binary subdivision$P_m:=\{ k2^{-m} \ : \ 0 \le k \le 2^m \},$we have$V(f_n;I)=V(f_n;P_m)$as soon as$n \geq m$. So for all$m$letting$n\to\infty$$$V(f;P_m)=\lim_{n\to\infty }V(f_n;P_m)=V(f_m;P_m)$$ and $$V(f;I)=\sup_{m\in\mathbb{N}}V(f;P_m)=\infty,$$ as we wished to show. 14 added 11 characters in body Your guess is indeed right. Following a similar idea gives you the Takagi or blancmange function. It is even quasi-Lipschitz (it has a modulus of continuity$\omega(t)=ct(|\log(t)|+1)$for a suitable constant$c>0$), thus it's Hoelder of any positive exponent less than 1. It is not even BV in any open interval, thus$L^p_{loc}$for no$p\geq1$. Rmk 1. The above example is for dimension 1: but of course it holds in any dimension a fortiori. Rmk 2. To get an example with a more classical flavor, actually a Weierstrass function, replace$s(x)$with$\cos(x)$. I'd say that the resulting Fourier series defines a function with the same features, by the same reasons (the function$\cos(x)$works better than$\sin(x),$in view of point 2 below.) Rmk 3. Once you know that the Weierstrass function$f(x):=\sum_{k=0}^\infty 2^{-k}\cos(2^k x)$is nowhere differentiable, you also have that it is BV on no open interval, for BV on an interval would imply differentiability a.e. there. However, for your needs it seems more direct just showing it has infinite variation on any interval. Details. 1. To prove that the Takagi function$f(x)$admits the above modulus of continuity, recall that that$f$is characterized as the fixed point of the affine contraction$T:C_b(\mathbb{R})\to C_b(\mathbb{R})$such that$(Tf)(x)=\frac{1}{2}f(2x)+s(x),$for all$x\in\mathbb{R}$, where$s(x)$is the distance function from the integers (a zig-zag piecewise 1-periodic function). Just find a$c$such that the subset of$C_b(\mathbb{R})$of functions that admit$\omega$as modulus of continuity is a$T$-invariant set. The latter subset is obviously closed and non-empty, so the fixed point is there. (The above illustrated a standard general technique to prove properties of objects found by means of the contraction principle). 2. Proving that$f$is not of bounded variation on$[0,1]$(hence in no open interval, due to the self-similarity encoded in the fixed point equation), requires a small computation on the partial sum$f_n$of the series defining$f$. Let $$f_n(x):=\sum_{k=0}^{n-1}\ 2^{-k} s(2^k x).$$ First note that the derivative of$f_n$only takes integer values, which of course come as a result of the sum of$n$terms$\pm 1$(with all the$2^n$possible signs). In particular, for any$n\in\mathbb{N}$the function$f_{2n}$has${2n}\choose {2n \choose n} $flat intervals of lenght$2^{-2n}$within the unit interval$I$, and has derivative larger than$2$in absolute value elsewhere in$I$. Thus, for the subsequent odd index$2n+1,$the function$f_{2n+1}$has${2n \choose n}$local maxima in$I$(located in the mid-points of the above intervals). Moreover, passing to$f_{2n+1}$each maximum point contributes to the increment of the total variation with$2^{-2n}$, while the total variation remains unchanged passing from$2n+1$to the next even index$2n+2$. The conclusion is that, for any$n$, the total variation of$f_n$on$I$is $$V(f_n;I)=\sum_{0\leq k < n/2}{2k\choose k}2^{-2k} =O\big(\sqrt{n}\big),$$ since by the classical asymptotics for the central binomial coefficient,${2k \choose k}=\frac{4^k}{\sqrt{\pi k}}(1+o(1)),\ k\to\infty.$So actually$V(f_n;I)$diverges. Yet this would not be sufficient to conclude that$V(f,I)=\infty,$as the total variation is only lower semicontinuous with respect to the uniform convergence. However, the discrete variation on a given subdivision$P:=\{t_0 < \dots < t_r \}$}$ $$V(f_n; P\ )=\sum_{i=0}^{r-1}|f_n(t_{i+1})-f(t_i)|$$ =\sum_{i=0}^{r-1}\ \big|f_n(t_{i+1})-f(t_i)\big|$$does of course pass to the limit under even pointwise convergence. Now the point is that, for the binary subdivision P_m:=\{ k2^{-m}k2^{-m} \ : \ 0\leq k\leq 0 \le k \le 2^m \}, we have V(f_n;I)=V(f_n;P_m) as soon as n \geq m. So for all m letting n\to\infty$$V(f;P_m)=\lim_{n\to\infty }V(f_n;P_m)=V(f_m;P_m)$$and$$V(f;I)=\sup_{m\in\mathbb{N}}V(f;P_m)=\infty,$$as we wished to show. 13 added 6 characters in body; edited body Your guess is indeed right. Following a similar idea gives you the Takagi or blancmange function. It is even quasi-Lipschitz (it has a modulus of continuity \omega(t)=ct(|\log(t)|+1) for a suitable constant c>0), thus it's Hoelder of any positive exponent less than 1. It is not even BV in any open interval, thus L^p_{loc} for no p\geq1. Rmk 1. The above example is for dimension 1: but of course it holds in any dimension a fortiori. Rmk 2. To get an example with a more classical flavor, actually a Weierstrass function, replace s(x) with \cos(x). I'd say that the resulting Fourier series defines a function with the same features, by the same reasons (the function \cos(x) works better than \sin(x), in view of point 2 below.) Rmk 3. Once you know that the Weierstrass function f(x):=\sum_{k=0}^\infty 2^{-k}\cos(2^k x) is nowhere differentiable, you also have that it is BV on no open interval, for BV on an interval would imply differentiability a.e. there. However, for your needs it seems more direct just showing it has infinite variation on any interval. Details. 1. To prove that the Takagi function f(x) admits the above modulus of continuity, recall that that f is characterized as the fixed point of the affine contraction T:C_b(\mathbb{R})\to C_b(\mathbb{R}) such that (Tf)(x)=\frac{1}{2}f(2x)+s(x), for all x\in\mathbb{R}, where s(x) is the distance function from the integers (a zig-zag piecewise 1-periodic function). Just find a c such that the subset of C_b(\mathbb{R}) of functions that admit \omega as modulus of continuity is a T-invariant set. The latter subset is obviously closed and non-empty, so the fixed point is there. (The above illustrated a standard general technique to prove properties of objects found by means of the contraction principle). 2. Proving that f is not of bounded variation on [0,1] (hence in no open interval, due to the self-similarity encoded in the fixed point equation), requires a small computation on the partial sum f_n of the series defining f. Let$$f_n(x):=\sum_{k=0}^{n-1}\ 2^{-k} s(2^k x).$$First note that the derivative of f_n only takes integer values, which of course come as a result of the sum of n terms \pm 1 (with all the 2^n possible signs). In particular, for any n\in\mathbb{N} the function f_{2n} has {2n}\choose n}  flat intervals of lenght 2^{-2n} within the unit interval I, and has derivative larger than 2 in absolute value elsewhere in I. Thus, for the subsequent odd index 2n+1, the function f_{2n+1} has {2n \choose n} local maxima in I (located in the mid-points of the above intervals). Moreover, passing to f_{2n+1} each maximum point contributes to the increment of the total variation with 2^{-2n}, while the total variation of remains unchanged passing from 2n+1 to the next even index 2n+2. The conclusion is that, for any n, the total variation of f_n on I is$$V(f_n;I)=\sum_{0\leq k < n/2}{2k\choose k}2^{-2k} =O\left(\sqrt(n)\rigth),$$O\big(\sqrt{n}\big),$$
since by the classical asymptotics for the central binomial coefficient, ${2k \choose k}=\frac{4^k}{\sqrt{\pi k}}(1+o(1)),\ k\to\infty,$ so k\to\infty.$So actually$V(f_n;I)$diverges. Yet this would not be sufficient to conclude that$V(f,I)=\infty,$as the total variation is only lower semicontinuous with respect to the uniform convergence. However, the discrete variation on a given subdivision$P:=\{t_0 < \dots < t_r \ }$$$V(f_n; P)=\sum_{i=0}{r}|f_n(t_{i+1})-f(t_i)|$$ P\ )=\sum_{i=0}^{r-1}|f_n(t_{i+1})-f(t_i)|$$does of course pass to the limit under even pointwise convergence. The Now the point is that, for the binary subdivision P_m:=\{ k2^{-m}\ : \ 0\leq k\leq 2^m \}, we have V(f_n;I)=V(f_n;P_m) as soon as n \geq m. So for all m letting n\to\infty$$V(f;P_m)=\lim_{n\to\infty }V(f_n;P_m)=V(f_m;P_m)$$and$$V(f;I)=\sup_{m\in\mathbb{N}}V(f;P_m)=\infty,$\$ as we wished to show.

12 added 9 characters in body
11 added 1370 characters in body
10 added 347 characters in body
9 deleted 6 characters in body; [made Community Wiki]
8 added 79 characters in body
7 added 246 characters in body
6 deleted 5 characters in body; added 6 characters in body