For each $n\in \Bbb N$, $n\ge1$ and for each $x\in \Bbb R$, consider the following operator: $T_n: L^2(\Bbb R) \to L^p(\Bbb R) $ defined as $$T_nf(x)=n^{3/4} \int_x^{x+1/n} f(t)\mathrm dt $$ The problem, as stated in the title, consists in finding for which $p\in[1,+\infty]$, $T_n$ is continuous. The case $p=+\infty$ is ok, I'm having trouble in the other cases. Since $T_n$ is linear, I have to prove that there is $K>0$ such that $\|T_nf\|_p\le K\|f\|_2$: $$\|T_nf \|_p=\left(\int_{-\infty}^{+\infty} \left|n^{3/4}\int_x^{x+1/n} f(t)\mathrm dt \right|^p\mathrm dx \right)^{1/p}=n^{3/4}\left(\int_{-\infty}^{+\infty} \left|\int_x^{x+1/n} f(t)\mathrm dt \right|^p\mathrm dx \right)^{1/p} $$ Since I must have $\|f \|_2$, I tried to apply Cauchy-Schwarz inequality in the inner integral but I get a divergent integral. I also tried to change coordinates (I was trying to do $p=1$ to see if I could generalize), noticing that I was integrating between the lines $t=x+1/n$ and $t=x$, but also in this case I get a divergent intergral.
I solved similar exercises using Minkowsi's integral inequality, but here I don't know if it's applicable.
This is true for all $p \geq 2$. I’ll first prove this for $p = 2$:
We have,
$$\begin{split} ||T_nf||_2^2 &= n^{\frac{3}{2}} \int_\mathbb{R} |\int_x^{x + 1/n} f(t) \, dt|^2 \, dx\\ &\leq n^{\frac{3}{2}} \cdot n^{-2} \int_\mathbb{R} (n\int_x^{x + 1/n} |f(t)| \, dt)^2 \, dx\\ &\leq n^{-\frac{1}{2}} \int_\mathbb{R} n\int_x^{x + 1/n} |f(t)|^2 \, dt \, dx\\ &= n^{\frac{1}{2}} \int_\mathbb{R} \int_{t - 1/n}^t |f(t)|^2 \, dx \, dt\\ &= n^{\frac{1}{2}} \cdot n^{-1} ||f||_2^2\\ &= n^{-\frac{1}{2}}||f||_2^2 \end{split}$$ Where the inequality on the third line uses Jensen’s inequality and the equality on the fourth line uses Fubini-Tonelli theorem. Thus, we have $||T_nf||_2 \leq n^{-\frac{1}{4}}||f||_2$.
You already mentioned the result for $p = +\infty$. For completeness, I’ll include a proof of that here as well: For any $x \in \mathbb{R}$,
$$\begin{split} |T_nf(x)| &= n^{\frac{3}{4}} |\int_x^{x + 1/n} f(t) \, dt|\\ &\leq n^{\frac{3}{4}} \cdot n^{-\frac{1}{2}} ||f||_2\\ &= n^{\frac{1}{4}} ||f||_2 \end{split}$$ Where the inequality on the second line uses Cauchy-Schwarz inequality. Hence, $||T_nf||_\infty \leq n^{\frac{1}{4}} ||f||_2$.
Now, for general $\infty > p \geq 2$. If $g \in L^2(\mathbb{R}) \cap L^\infty(\mathbb{R})$, then,
$$\begin{split} ||g||_p^p &= \int_\mathbb{R} |g(x)|^p \, dx\\ &= \int_\mathbb{R} |g(x)|^2 |g(x)|^{p - 2} \, dx\\ &\leq ||g||_\infty^{p - 2} ||g||_2^2 \end{split}$$
Thus,
$$\begin{split} ||T_nf||_p^p &\leq ||T_nf||_\infty^{p - 2} ||T_nf||_2^2\\ &\leq n^{\frac{p - 2}{4}} ||f||_2^{p - 2} \cdot n^{-\frac{1}{2}} ||f||_2^2\\ &= n^{\frac{p - 4}{4}} ||f||_2^p \end{split}$$
So $||T_nf||_p \leq n^{\frac{p - 4}{4p}} ||f||_2$ for all $\infty > p \geq 2$. If we interpret $n^{\frac{p - 4}{4p}}$ as $n^{\frac{1}{4}}$ when $p = +\infty$, then the same holds when $p = +\infty$ as well, as we have already seen.
The following shows that the map is not bounded when $1 \leq p < 2$. In fact, the range is not even contained in $L^p(\mathbb{R})$ when $1 \leq p < 2$.
For any $0 < \epsilon < 1$, let $f(t) = t^{-\frac{1 + \epsilon}{2}} 1_{\{t > 1\}}$. One easily checks that $f \in L^2(\mathbb{R})$. When $x > 1$, we have,
$$\begin{split} T_nf(x) &= n^{\frac{3}{4}} \cdot \frac{2}{1 - \epsilon}[(x+\frac{1}{n})^{\frac{1-\epsilon}{2}} - x^{\frac{1-\epsilon}{2}}]\\ &= \frac{2n^{\frac{3}{4}}}{1 - \epsilon}x^{\frac{1-\epsilon}{2}}[(1+\frac{1}{nx})^{\frac{1-\epsilon}{2}} - 1] \end{split}$$
By the generalized binomial theorem, as $x \to \infty$,
$$(1+\frac{1}{nx})^{\frac{1-\epsilon}{2}} - 1 \approx \frac{1-\epsilon}{2} \cdot \frac{1}{nx}$$
So, as $x \to \infty$,
$$\begin{split} T_nf(x) &= \frac{2n^{\frac{3}{4}}}{1 - \epsilon}x^{\frac{1-\epsilon}{2}}[(1+\frac{1}{nx})^{\frac{1-\epsilon}{2}} - 1]\\ &\approx \frac{2n^{\frac{3}{4}}}{1 - \epsilon}x^{\frac{1-\epsilon}{2}} \cdot \frac{1-\epsilon}{2} \cdot \frac{1}{nx}\\ &= n^{-\frac{1}{4}}x^{-\frac{1+\epsilon}{2}} \end{split}$$
But then, whenever $1 \leq p \leq \frac{2}{1+\epsilon}$, because we then have $-\frac{p+p\epsilon}{2} \geq -1$, so,
$$\begin{split} ||T_nf||_p^p &\geq ||1_{(1, \infty)}T_nf||_p^p\\ &\approx \int_1^\infty n^{-\frac{p}{4}}x^{-\frac{p+p\epsilon}{2}} \, dx\\ &= +\infty \end{split}$$
As $\lim_{\epsilon \to 0} \frac{2}{1+\epsilon} = 2$, this shows that the range of $T_n$ is not contained in $L^p(\mathbb{R})$ for all $n$ and $1 \leq p < 2$.