This is not an answer, but rather a working-out of the choice of $f$ I mentioned in the question.

Let $f = (\log y)^{-k} (\Delta_{1/y}^\cdot)^k g,$ where $g(x) = \log^k(1/x)$ for $0<x\leq 1$ and $g(x)=0$ for $x>1$. It is easy to see that $f(x)=1$ for $0\leq x\leq 1$ and $f(x)=0$ for $x\geq y^k$. Hence $$|f-1_{[0,1]}|_1 = \sum_{j=0}^{k-1} \int_{y^j}^{y^{j+1}} f(x) dx = \dotsc$$

Wait, the long answer I had lovingly crafted just got erased when I closed a window. (I had been working off-line.) The final conclusion was that the quantity we wish to minimize ends up being roughly $(k/2) (2/T)^{(k-1)/k}$, that we are best off taking $k = (\log T)/2$, and that the bound then is about $$\frac{e \log T}{T}.$$

Can one do much better?

(If one drops the condition $f(x)=1$, then of course one can do better, by using $f(y^{k/2} x)$ instead of $f$. Then one saves a constant of at least $2$. I can work this out in lieu of my vanished answer, if people are interested.)

This is not an answer, but rather a working-out of the choice of $f$ I mentioned in the question.

Let $f = \frac{(\log y)^{-k}}{k!} (\Delta_{1/y}^\cdot)^k g,$ where $g(x) = \log^k(1/x)$ for $0<x\leq 1$ and $g(x)=0$ for $x>1$. It is easy to see that $f(x)=1$ for $0\leq x\leq 1$ and $f(x)=0$ for $x\geq y^k$. Hence $$|f-1_{[0,1]}|_1 = \sum_{j=0}^{k-1} \int_{y^j}^{y^{j+1}} f(x) dx = \dotsc$$

Wait, the long answer I had lovingly crafted just got erased when I closed a window. (I had been working off-line.) The final conclusion was that the quantity we wish to minimize ends up being roughly $(k/2) (2/T)^{(k-1)/k}$, that we are best off taking $k = (\log T)/2$, and that the bound then is about $$\frac{e \log T}{T}.$$

Can one do much better?

(If one drops the condition $f(x)=1$, then of course one can do better, by using $f(y^{k/2} x)$ instead of $f$. Then one saves a constant of at least $2$. I can work this out in lieu of my vanished answer, if people are interested.)

This is not an answer, but rather a working-out of the choice of $f$ I mentioned in the question.

Let $f = (\log y)^{-k} (\Delta_{1/y}^\cdot)^k g,$ where $g(x) = \log^k(1/x)$ for $0<x\leq 1$ and $g(x)=0$ for $x>1$. It is easy to see that $f(x)=1$ for $0\leq x\leq 1$ and $f(x)=0$ for $x\geq y^k$. Hence $$|f-1_{[0,1]}|_1 = \sum_{j=0}^{k-1} \int_{y^j}^{y^{j+1}} f(x) dx = \dotsc$$

Wait, the long answer I had lovingly crafted just got erased when I closed a window. (I had been working off-line.) The final conclusion was that the quantity we wish to minimize ends up being roughly $(k/2) (2/T)^{(k-1)/k}$, that we are best off taking $k = (\log T)/2$, and that the bound then is about $$\frac{e \log T}{T}.$$

Can one do much better?

This is not an answer, but rather a working-out of the choice of $f$ I mentioned in the question.

Let $f = (\log y)^{-k} (\Delta_{1/y}^\cdot)^k g,$ where $g(x) = \log^k(1/x)$ for $0<x\leq 1$ and $g(x)=0$ for $x>1$. It is easy to see that $f(x)=1$ for $0\leq x\leq 1$ and $f(x)=0$ for $x\geq y^k$. Hence $$|f-1_{[0,1]}|_1 = \sum_{j=0}^{k-1} \int_{y^j}^{y^{j+1}} f(x) dx = \dotsc$$

Wait, the long answer I had lovingly crafted just got erased when I closed a window. (I had been working off-line.) The final conclusion was that the quantity we wish to minimize ends up being roughly $(k/2) (2/T)^{(k-1)/k}$, that we are best off taking $k = (\log T)/2$, and that the bound then is about $$\frac{e \log T}{T}.$$

Can one do much better?

(If one drops the condition $f(x)=1$, then of course one can do better, by using $f(y^{k/2} x)$ instead of $f$. Then one saves a constant of at least $2$. I can work this out in lieu of my vanished answer, if people are interested.)