# Convolution - Hölder inequality

I wonder if you guys can help me out with a question(not homework).

I have $\phi(x)=\int_\mathbb{R} |f(t)g(x-t)|dt$ where $f \in L^1(\mathbb{R})$ and $g \in L^p(\mathbb{R})$ and p and p' are conjugate exponents ($p \in (1,\infty).$ I want to show that $\phi(x) \leq \left( \int_\mathbb{R} |f(t)| dt \right)^{1/p'} \left( \int_{\mathbb{R}} |f(t)g(x-t)|^p dt \right)^{1/p}$.

The correction says that I should put $|f(t)g(x-t)|=|f(t)|^{1/p'}|f(t)|^{1/p}|g(x-t)|$

Is the argument here that $f\in L^1(\mathbb{R})$ implies $f^{1/p} \in L^{p'}$ and $f^{1/{p'}} \in L^{p'}$?

Can I use Jensens inequality to show this? $\int_\mathbb{R} |f(t)|dt < \infty \Rightarrow \left( \int_\mathbb{R} |f(t)|dt \right)^{1/p} \leq \int_\mathbb{R} |f(t)|^{1/p}dt$

I'm incertain of how I can treat these kids of integral inequalities.

• I understand nearly nothing. can you write the Holder inequality ? – reuns May 11 '16 at 14:52
• $f \in L^p(\mathbb{R}) g \in L^{p'}(\mathbb{R})$ where $1/p + 1/{p'} =1$ $\Rightarrow \parallel fg \parallel_1 \leq \parallel f \parallel_p \parallel g \parallel_{p'}$. What is it that I need to clarify? – user202542 May 11 '16 at 15:47
• @user202542 Are you sure about the inequality you want proved? I would believe it reads $$\phi(x) \leq \left( \int_\mathbb{R} |f(t)| dt \right)^{1/p'} \left( \int_{\mathbb{R}} |g(t)|^p|f(t)| dt \right)^{1/p}$$ instead. – zuggg May 13 '16 at 12:13
• I think it's right! It's from an old exam and this is what the correction says... How do you prove the one you propose? – user202542 May 15 '16 at 8:06

As you wrote in a comment we have the Hölder inequality:

If $h\in L^p(\mathbb{R})$ and $k\in L^{p'}(\mathbb{R})$, where $1/p+1/p'=1$ and $1\leq p,\,p'\leq +\infty$ then $$\int_\mathbb{R} |h(x)k(x)|dx\leq \left(\int_\mathbb{R} |h(x)|^pdx\right)^{1/p}\left(\int_\mathbb{R} |k(x)|^{p'}dx\right)^{1/p'}$$

In our case we look at $f\in L^1(\mathbb{R})$ and $g\in L^{p}(\mathbb{R})$, and we wish to estimate their convolution (in a point-wise sense) $$f*g(t) =\int f(x) g(t-x)\,dx.$$ However, since $1/p+1/1\ne1$ something has to be done in order to achieve a proper estimate.

One thing we always can do is to find the exponent $p'$ appearing in Hölder's inequality that correspond to $p$ (the so called conjugate exponent of $p$): it is the number $p'=p/(p-1)$. With this choice we have $$|f(x)|= |f(x)|^1= |f(x)|^{1/p+ 1/p'}= |f(x)|^{1/p}\cdot|f(x)|^{1/p'}$$ which leads to \begin{eqnarray*} \int|f(x)||g(t-x)|dx &=& \int |f(x)|^{1/p'}\cdot|f(x)|^{1/p}|g(t-x)|dx\\ &\leq&\left(\int |f(x)|^{p'/p'}\right)^{1/p'}\left(\int |f(x)|^{p/p}|g(t-x)|^p\,dx\right)^{1/p}\\ &=&\|f\|_{L^1}^{1/p'}\left(\int |f(x)||g(t-x)|^p\,dx\right)^{1/p} \end{eqnarray*} by Hölder's inequality - this is bounded because both $f$ and $|g|^p$ belongs to $L^1$, and $L^1$ is closed under convolution.

For your last query regarding validity of $\left(\int |f(x)| dx\right)^{1/p} \leq \int |f(x)|^{1/p}$ - This is false. It is true for the counting measure on $\mathbb{Z}$, but not on $\mathbb{R}$.

Counter example: Define $f$ as piece-wise constant and unbounded as follow. For each integer $n >0$ we write $f(x) = n$ if $x\in(n,n+1/n^2)$, elsewhere we set $f(x)=0$, then $$\int f(x) dx = \sum_{n>0} \int_n^{n+1/n^2}n\, dx = \sum_{n>0} n\cdot\frac{1}{n^2} = \infty$$ while $$\int f(x) dx = \sum_{n>0} \int_n^{n+1/n^2}n\, dx = \sum_{n>0} n\cdot\frac{1}{n^2} = \infty$$

while $$\int |f(x)|^{1/p} dx =\sum_{n>0} \int_n^{n+1/n^2}n^{1/p}\, dx = \sum_{n>0} n^{1/p}\cdot\frac{1}{n^2} < \infty$$ because $2-1/p>1$ for $p>1$.