Problem: Let $\pi_k\varphi(t)$ be the $L_2-$projection of $\varphi$ into piecewise constants, i.e
$$\int_{I_j}\pi_k\varphi(t)=\int_{I_j}\varphi(t).\tag 1$$
Show that for a subinterval $I_j=(t_{j-1}, t_j)$, with $t_j=jk$ and $k$ being a positive constant
$$\int_{I_j}|\varphi(s)-\pi_k\varphi(s)| \ ds\leq k\int_{I_j}|\dot{\varphi}(s)| \ ds, \quad \dot{\varphi}(t)=\frac{d\varphi}{dt}. \tag2$$
Question 1:
1) Why does $(1)$ hold? How do I intuitively see it?
Question 2 & 3: In the solution suggestion they write
2) Why is that first assumption crucial? Can't we project $\varphi$ without it crossing path with $\pi_k\varphi$?
3) I don't see how that last equation follows. The LHS just gives us the vertical distance between the red and blue line. If we evaluate the integral we get
$$\varphi(t)-\pi_k\varphi(t)=\varphi(t)-\varphi(\tilde{t}) \ \Longleftrightarrow \pi_k\varphi(t)=\varphi(\tilde{t}), \tag 3$$
Which is clearly not true. What am I missing?
Let $\mathcal F:= \{ \sum_{n\in \mathbb Z} c_n \chi_{I_n}: c_n \in \mathbb R, I_n = [(n-1)k,nk), \sum_{n\in \mathbb Z} c_n^2<\infty\}$ denote the inner product space of the piecewise constant functions you mentioned. In $L^2(\mathbb R)$ the inner product is defined by $\langle f ,g \rangle = \int_{\mathbb R} f(x)g(x)dx .$ It is easy to show that the set $\{\frac{1}{\sqrt k} \chi_{I_n} , n\in\mathbb Z \}$ is an orthonormal basis of $\mathcal F$, since $\int \frac{1}{\sqrt k}\chi_{I_n}(x) \frac{1}{\sqrt k} \chi_{I_m}(x)dx = \delta_{nm}. $ Okay so the othogonal projection of $\varphi\in L^2$ is given by $$ P_{\mathcal F} \varphi(x) = \sum_{n\in \mathbb Z} \langle \varphi,v_n\rangle v_n(x) = \sum_{n\in\mathbb Z} \frac{1}{k}\int_{I_n} \varphi(t)dt\cdot \chi_{I_n}(x)= \sum_{n\in\mathbb Z} \pi_k\varphi(x), $$ where $v_n(x):=\frac{1}{\sqrt k} \chi_{I_n}(x)$. Now fix $x\in I_n$ for some $n$. Due the integral mean value theorem $$ \frac1k\int_{I_n} \varphi(x)dx = \varphi(\xi) $$ for some $\xi\in I_n$. We also know that $$ \frac1k \int_{I_n} \varphi(x)dx = \pi_k\varphi(t) \quad \mbox{ for any } t \in I_n $$ hence for $\eta>\xi$ $$ \varphi(\eta)-\pi_k\varphi(\eta) =\varphi(\eta)-\varphi(\xi)= \int_{\xi}^\eta \varphi'(t)dt \leq \int_{\xi}^\eta |\varphi'(t)|dt $$ from which follows $$ |\varphi(\eta)-\pi_k\varphi|\le \int_{\xi}^\eta|\varphi'(t)|dt .$$ If $\eta<\xi$ then $$ \pi_k\varphi- \varphi(\eta) =\int_{\eta}^\xi\varphi'(t)dt $$ from which we get similarly $$ |\varphi(\eta)-\pi_k\varphi| \leq \int_{\eta}^\xi |\varphi'(t)|dt $$ Now we have $$ |\varphi(\eta)-\pi_k\varphi|\leq \int_{\min(\xi,\eta)}^{\max(\xi,\eta)}|\varphi'(t)|dt \le \int_{I_n} |\varphi'(t)|dt \quad \mbox{ for any } \eta \in I_n $$ Now integrating this w.r.t. to $\eta$ yields $$ \int_{I_n} |\varphi(\eta)-\pi_k\varphi|d\eta\leq k\int_{I_n}|\varphi'(t)|dt $$