I'm stuck in the exercise 115 in this book. Here $f\approx g$ means $C_1 g\le f\le C_2 g$ for some positive constants $C_1,C_2$ and $B_1(t)=t-\lfloor t\rfloor-1/2$.
It is about the Van der Corput's method and the squarefree numbers in short interval. First, it requires to show that (I didn't do this but seems to need the Van der Corput's theorem since it's pretty similar)
Let $R\ge 2$ be an integer and $N\ge 1$ also integer. Let $I$ be the interval in $[N+1,N]$. Show that if $f\in C^R(I,\mathbb R)$ and if there exists $F>0$ such that $|f^{(r)}(t)|\approx FN^{-r}$ for all $1\le r\le R, x\in I$, then $$\sum_{n\in I}B_1(f(n))\ll (FN^{-R})^{1/(2^R-1)}N+F^{-1}N.$$
And then there is another exercise to show that if we let $$S_4:=\sum_{\substack{x-y<d^2m\le x\\d>D}}1,$$ for $\sqrt{y}\le D\le \sqrt{x}$, $S_4\ll xD^{-2}+yD^{-1}.$ It seems to me that I need the above lemma but as I tried it's not the same as the exercise.
Here is what I tried (I even use the above with $R=1$ since it is near the problem but the lemma may not allow and sadly fail too).
First, we have $$S_4=\sum_{d>D}\left(\left\lfloor \dfrac{x}{d^2}\right\rfloor-\left\lfloor \dfrac{x-y}{d^2}\right\rfloor\right)=\sum_{\sqrt{x}\ge d>D}\dfrac{y}{d^2}-\sum_{\sqrt{x}\ge d>D}\left(B_1\left(\dfrac{x}{d^2}\right)-B_1\left(\dfrac{x-d}{d^2}\right)\right)=: S_{4,1}-S_{4,2},$$ and $S_{4,1}\ll yD^{-1}$ so I assume that I need to show that $$\sum_{\sqrt{x}\ge d>D} B_1\left(\dfrac{x}{d^2}\right)\ll xD^{-2}.$$
Then, even I allow myself the above with $R=1$ with $I_k=[2^{k-1}D, 2^kD]$ (for $k=1,2\dots, K\approx \log(\sqrt{x}/D)$) and $f(t)=x/t^2$ so I can choose $F=x/D^2$. I have from the (miracle) lemma, $$\sum_{\sqrt{x}\ge d>D}B_1(f(d))\ll \sum_{k\le K}\sum_{d\in I_k}B_1(f(d))\ll \dfrac{x}{D^2}+\dfrac{D^3}{x},$$ by assuming gratefully that $D=x^{\delta}$ for some $\delta>0$ (so $K\approx \text{const}$). So I have $D^3/x$ as excession. It's neither true that $D^3/x\ll x/D^2$ nor $\ll y/D$ by just the definition of $D$. However, it is true that when it's used later ($D=X^{3/8}$) and this term get absorbed but I don't think it is justifyable since this should hold for all $x^\delta\le D\le x^{1/2}$? Also, I used the version not existing and I don't know how to show it the other way.
Please help me on this, I would be grateful if the exercise is true already since I need the similar result further.