Jtdylktuy

We're given the function

$$f(x) = begin{cases} 2 sin(x) + 3, & x> 0 \
-2 sin(x) + c, & x leq 0
end{cases} \$$

and asked to find the values of $c$ for which $f in W^{1,p}_{mathrm{loc}}$.

To start, I've found the distributional derivative of $f$ to be

$$f'(x) = begin{cases} 2 cos(x), & x> 0 \
-2 cos(x), & x leq 0
end{cases} \$$

and I want to find values of $c$ for which $leftlVert frightrVert_{L^p(K)} < infty$ where $K$ is any closed ball in $mathbb{R}$. But I'm stuck on that computation.

Your claimed distributional derivative is not correct in general. Indeed you may pick up a extra Dirac delta term arising from the fact that,
$$frac{mathrm{d}}{mathrm{d}t}, 1_{(-infty,0]} = delta_0.$$

The following rules out all but possibly one value of $c.$

Claim: If $f in W^{1,p}_{mathrm{loc}}(mathbb R) subset W^{1,1}_{mathrm{loc}}(mathbb R),$ then $f$ is continuous (more precisely it admits a continuous representative).

Proof of claim: Note that if $g in C^{infty}_c(mathbb R),$ then by the fundamental theorem of calculus and $a < x < b,$
begin{align*}
|g(x)| &= left|g(a)+int_{a}^x g'(t) ,mathrm{d}t right| \
&leq frac1{b-a}lVert g rVert_{L^1([a,b])} + frac1{b-a}int_a^b |g(a)-g(x)|,mathrm{d}t + lVert g' rVert_{L^1((a,b))} \
&leq C_{a,b} lVert g rVert_{W^{1,1}([a,b])}.
end{align*}
So $lVert g rVert_{L^{infty}([a,b])} leq C lVert g rVert_{W^{1,1}([a,b])}$ for all $g in C^{infty}([a,b])$ and $[a,b] subset mathbb R.$ By approximating $f$ in $W^{1,1}(mathbb R)$ via mollification, we get a sequence of smooth functions which converges locally uniformly to $f$ by the above estimate. Hence $f$ is continuous in $mathbb R.$ $square$

Now your $f$ as defined is continuous if and only if $c=3.$ So we get $f not in W^{1,p}_{mathrm{loc}}(mathbb R)$ if $c neq 3.$

Now what happens if $c=3$? We first need to show that $f'$ is what you claim, so let
$$g(x) = begin{cases} 2 cos(x) &text{if } x> 0 \
-2 cos(x) &text{if } x leq 0
end{cases} \$$
be the candidate for $f'$ and fix an arbitrary $varphi in C^{infty}_c(mathbb R).$ Then as $f$ is continuous on $mathbb R$ and $C^1$ with derivative $g$ on $mathbb R setminus {0}$, integrating by parts on $(-infty,0)$ and $(0,infty)$ we get,
begin{align*}
int_{-infty}^{infty} varphi'(t) f(t) ,mathrm{d}t &= int_{-infty}^0 varphi'(t) f(t) + int_0^{infty} varphi'(t) f(t),mathrm{d}t \
&= left(varphi'(0)f(0) - int_{-infty}^0 varphi(t)g(t) ,mathrm{d}tright) + left(-varphi'(0)f(0) - int_0^{infty} varphi(t)g(t) ,mathrm{d}tright) \
&= - int_{-infty}^{infty} varphi(t)g(t).
end{align*}
Thus $f$ is weakly differentiable with weak derivative $f' = g.$ Note that we needed continuity of $f$ at $0$ to integrate by parts.

Finally as both $f$ and $f'$ are bounded, we get $f, f' in L^{infty}(mathbb R)$ and hence $f,f' in L^p_{mathrm{loc}}(mathbb R)$ for all $p$ by Hölder. Hence $f in W^{1,p}_{mathrm{loc}}(mathbb R)$ for all $1 leq p leq infty$ when $c=3.$