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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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    No. As you observe, the function $f$ in your question belongs to $L^1 [0,1]$ but not to $L^\infty [0,1]$.2011-10-04
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    The statement you refer to on convolutions is a special case of [Young's equality](http://en.wikipedia.org/wiki/Young_inequality#Young.27s_inequality_for_convolutions). Its proof is, indeed, not _so_ simple.2011-10-04
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    By the way, you are computing the wrong thing. $\int|f(t-s)\phi_n(s)|^p~\mathrm{d}s$ is **not** what you should be computing. What you should be computing is $$\int \left|\int f(t-s)\phi_n(s)~\mathrm{d}s\right|^p\mathrm{d}t$$ which is emphatically different. The usual way of proving Young's inequality is via interpolation: the case where $p= \infty$ is trivial, and the case where $p = 1$ is not too hard with a suitable version of Fubini.2011-10-04
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    @Willie Wong: Well, this is one of the easy cases of Young's *inequality* - and I see nothing wrong in the $L^\infty$-estimate of the OP (so don't scare him/her) :)2011-10-04
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    @AD. I take "summability kernel" to mean that $\phi_n\in L_1(\mathbb{T})$.2011-10-04
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    $L^p(X) \subseteq L^\infty(X)$ for a certain measure space provided $X$ does not admit arbitrarily small sets: sets with positive arbitrarily small measure.2011-10-04
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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