This is a problem from Rene Schilling's Measures, Integrals and Martingales which shows that the associativity of the convolution may fail.
Define $\phi: \mathbb{R} \to \mathbb{R}$ by $\phi(x):= (1-\cos x) \mathbb{1}_{[0,2\pi)}(x),$ let $u(x):=1,$ $v(x):= \phi'(x)$ and $w(x):= \int_{(-\infty,x)}\phi(t)dt.$ Then $v \star w(x) = \phi \star \phi (x) >0 $ for all $x \in(0,4\pi)$.
Below is the solution to this problem. We should have $\int v(x-y)w(y)dy$. From the definition of $v$, what is the derivative with respect to? I am not sure what $'$ specifically refers to here. Also, how do we get the equivalence of the second and third integrals using integration by parts? I would greatly appreciate any help.
By definition the function $\phi$ maps from $\mathbb{R}$ to $\mathbb{R}$. Hence there is no ambiguity in taking derivatives: there is only one variable and one direction.
Furthermore the notation $\phi'(x)$ is an abbreviation for $$\bigg(\frac{\partial \phi}{\partial t}(t)\bigg) \bigg|_{t=x}.$$
First the function gets differentiated, then you plug in the point you are evaluating at. Under the integral a different point of view is taken: First the function $x \mapsto x-y$ gets plugged into $\phi$ and then the operator $\frac{\partial }{\partial x}$ differentiates this composition. The inner derivative is 1, so we just reproduce our old result. Using the operator $\frac{\partial }{\partial y}$ instead yields an additional minus sign, which is the second step in the above computation.
With that in mind we can now look at the entire thing. Plugging in all the definitions immediately makes it more confusing than it has to be.
Only using the definition of the convolution product $\ast$ yields $$\big(v \ast w\big)(x) = \int_\mathbb{R} v(x-y) w(y) d y = \int_\mathbb{R} \phi'(x-y) w(y) d y = - \int_\mathbb{R} \bigg(\frac{\partial}{\partial y} \phi(x-y)\bigg) w(y) d y.$$
Now we can clearly integrate by parts. Because the support of $\phi$ is compact, the boundary term just vanishes. Hence we have by the fundamental theorem of calculus $$ \big(v \ast w\big)(x) = \int_\mathbb{R} \phi(x-y) w'(y) d y = \int_\mathbb{R} \phi(x-y) \phi(t) \big|_{t=-\infty}^{t=y} d y = \int_\mathbb{R} \phi(x-y) \phi(y) d y = \big( \phi \ast \phi\big)(x).$$ Looking at the explicit form of $\phi$ then yields the desired inequality on the nose, as described in your screenshot.