Show $\lim_{j\to\infty}j^{(2p-2)/p} \left| \int_{1/j+1}^{1/j}f(x)\,dx\right| =0$ Assuming for $1<p<\infty$ that $f \in L^p([0,1])$ and denote $\| f\|_p = M< \infty$, here's what I have so far:
$$\lim_{j\to\infty}j^{(2p-2)/p}\left|\int_{1/j+1}^{1/j}f(x)\,dx \right| = \lim_{j\to\infty}j^{(2p-2)/p} \left|\int_{[0,1]}\mathbf{1}_{[1/j+1,1/j]}f(x)\,dx\right| \leq \lim_{j\to\infty}j^{(2p-2)/p}\int_{[0,1]}|\mathbf{1}_{[1/j+1,1/j]}f(x)|\,dx $$
Now that integral is just $\|\mathbf{1}_{[1/j+1,1/j]}f(x) \|_1$ so by Holder's inequality:
$$\|\mathbf{1}_{[1/j+1,1/j]}f(x) \|_1 \leq \|\mathbf{1}_{[1/j+1,1/j]} \|_{\frac{p}{p-1}}\|f(x)\|_p = M\bigg(\int_{[0,1]}|\mathbf{1}_{[1/j+1,1/j]}|^{\frac{p}{p-1}}\,dx \bigg)^{\frac{p-1}{p}}=M\left[\frac{1}{j(j+1)}\right]^{\frac{p-1}{p}} $$
And then substituting into the limit,
$$\lim_{j\to\infty}j^{(2p-2)/p}\int_{[0,1]}|\mathbf{1}_{[1/j+1,1/j]}f(x)|\,dx \leq \lim_{j\to\infty}j^{\frac{2(p-1)}{p}}\left[\frac{1}{j(j+1)}\right]^{\frac{p-1}{p}}M = \lim_{j\to\infty}M \left[\frac{j^2}{j(j+1)}\right]^{\frac{p}{p-1}}=M$$
Have I messed up somewhere with my exponents? Or is this just wrong entirely? I was told to use Holder's as a hint.
 A: Set $g_{j}(x) = j^{2/q}\cdot1_{[1/(j+1),1/j]}(x)$, where $q=p/(p-1)$ denotes the conjugate Hölder exponent of $1<p<\infty$ and notice that your problem is equivalent to proving that sequence $\left\{g_{j}\right\}_{j=1}^{\infty}$ converges weakly to $0$ in $L^{q}([0,1])$. The proof can be carried out in various ways, but the simplest one that comes to mind is to use the fact that compactly supported $L^{p}$-functions on $(0,1)$ are dense in $L^{p}([0,1])$. To see this, consider $f_{n}= 1_{[1/n, 1/(n+1)]}f$, for $f\in L^{p}([0,1])$ and use the dominated convergence theorem to show that $f_{n}\rightarrow f$ in $L^{p}([0,1)]$.
To this end, let $f$ be an arbitrary function in $L^{p}([0,1])$ and $\varepsilon >0$. By density, we can pick a $f_{\varepsilon}\in L^{p}([0,1])$ with compact support on $(0,1)$, such that $\| f-f_{\varepsilon} \|_{p}< \varepsilon$, and choose a large integer $N_{\varepsilon}>0$ with the property that the interval $[0,1/N_{\varepsilon}]$ does not intersect the support of $f_{\varepsilon}$. As you already observed, it is straightforward to check that $\| g_{j} \|_{q} \leq 1$, for all $j\geq 1$. Moreover, it follows that for all $j> N_{\varepsilon}$, we have
$$\int_{0}^{1}f_{\varepsilon}g_{j}dx = 0.
$$
Now combining this and applying Hölder's inequality, we get
$$ \lvert \int_{0}^{1}fg_{j}dx \rvert = \lvert \int_{0}^{1}(f-f_{\varepsilon})g_{j} dx \rvert \leq \| f-f_{\varepsilon} \|_{p}  < \varepsilon \qquad ,\, \forall j \geq N_{\varepsilon}. 
$$
This proves the desired claim.
