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I was about to post yet another question about a comment in my notes. But I think my key point in misunderstanding jumps in proofs is how the $L_{p}$ spaces are nested in eachother.

The example that always makes me question this is the function $f(x) = \frac{1}{\sqrt{x}}$ which is not bounded on $[0,1]$ yet is integrable.

The following remark gives rise to my question:

If $\{\phi_{n}\}$ is a summability kernel and $f\in L_{p}(\mathbb{T})$, then we immediately have $f*\phi_{n}\in L_{p}(\mathbb{T})$.

When I begin to compute $\int_{-\pi}^{\pi}|f(t-s)\phi_{n}(s)|^pds$ I get this

\begin{eqnarray*} \int_{-\pi}^{\pi}|f(t-s)\phi_{n}(s)|^pds &=& \int_{-\pi}^{\pi}|f(t-s)|^{p}\cdot|\phi_{n}(s)|^pds \end{eqnarray*}

if I had $\phi_{n}\in L_{\infty}(\mathbb{T})$, then I could immediately get $ \begin{eqnarray*} \int_{-\pi}^{\pi}|f(t-s)|^{p}\cdot|\phi_{n}(s)|^{p}ds &\leq& \int_{-\pi}^{\pi}|f(t-s)|^{p}\cdot||\phi_{n}||_{\infty}^{p}ds\\ &\leq& ||\phi_{n}||_{\infty}^{p}\int_{-\pi}^{\pi}|f(t-s)|^{p}ds\\ &<& \infty \end{eqnarray*}$

This would make the answer yes. But the example I mentioned above makes me think this might not be true; and thus the answer to my question would be non-trivial.

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    @GEdgar: An example of that is the counting measure on $\mathbb{Z}$.2011-10-04

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Your example can be used to show that $L^p(0,1)$ is not in $L^\infty(0,1)$.

Take any $p<\infty$ and put $f(x)= x^{-1/2p}$, then $\int_0^1 |f(x)|^pdx=\int_0^1 (x^{-1/2p})^pdx= \int_0^1x^{-1/2}dx=2$ but $f$ is unbounded at $x=0$.

It is actually the other way around. If $p then $L^q(0,1)\subset L^p(0,1)$ -- as follows from Hölder's inequality with parameters $p_1=q/p$ and $p_2=q/(q-p)$ (you should estimate the norm of $|f(x)|^p=1\cdot |f(x)|^p$).