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I write the exact statement of the problem:

Show that for any $g \in L^1$ and $f ∈ L^p(\mathbb{R})$, p $\in (1, \infty)$, the integral for f ∗g converges absolutely almost everywhere and that $∥f ∗ g∥_p ≤ ∥g∥_1∥f∥_p$.

I've tried to show that the convolution is well defined for f,g $\in L^1$ in this way:

$$\int_{\mathbb{R^d}}f*gdx=\int_{\mathbb{R^d}}\int_{\mathbb{R^d}}f(x-y)g(y)dydx\le\int_{\mathbb{R^d}}\int_{\mathbb{R^d}}|f(x-y)||g(y)|dydx$$

I've used Tonelli's theorem since the function is non-negative

$$\int_{\mathbb{R^d}}|g(y)|\int_{\mathbb{R^d}}|f(x-y)|dxdy=∥f∥_1∥g∥_1$$

Now I've thought that since $L^1 \subset L^p$ this is true also for f $\in L^p$ (can I do it? )


I've tried with some colleagues something with Hölder's inequality. I've chosen a q $\in$ (1,$\infty$) such that $\frac{1}{p}+\frac{1}{q}=1$

$$∥f ∗ g∥_p^p=\int_{\mathbb{R^d}}|\int_{\mathbb{R^d}}f(x-y)g(y)dy \space |^pdx=\int_{\mathbb{R^d}}\space |\int_{\mathbb{R^d}}f(x-y)g(y)^{\frac1p}g(y)^{\frac1q}dy\space |^pdx$$

$$\le\int_{\mathbb{R^d}}\space \int_{\mathbb{R^d}}|f(x-y)g(y)^{\frac1p}g(y)^{\frac1q} \space |^pdy\space dx=\int_{\mathbb{R^d}}\space \int_{\mathbb{R^d}}|f(x-y)^pg(y)g(y)^{\frac{p}{q}}|\space dy\space dx$$

$$\int_{\mathbb{R^d}}\space \int_{\mathbb{R^d}}|f(x-y)^pg(y)g(y)^{\frac{p}{q}}| \space dy\space dx=\int_{\mathbb{R^d}}\space \int_{\mathbb{R^d}}|f(x-y)^pg(y)^p| \space dy\space dx$$

$$= \int_{\mathbb{R^d}}\space \int_{\mathbb{R^d}}|f(x-y)^pg(y)^p| \space dx\space dy=∥g∥_1^p∥f∥_p^p \implies ∥f ∗ g∥_p ≤ ∥g∥_1∥f∥_p$$

This works is it not?

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    $\begingroup$ $L^1(\mathbb{R})$ isn't contained in $L^p(\mathbb{R})$. You need a bit of Hölder [there probably are other ways too] to reach the conclusion. $\endgroup$ Jan 16 '14 at 16:47
  • $\begingroup$ The problem with inequalities such as $L^p(B) \subset L^1(B)$ for $p>1$ is that the proof requires $B$ to be of finite measure. In that case, $p<q$ implies $L^q(B)\subset L^p(B)$ (by an adequate use of Hölder's inequality). Just write $|f(x)|=(|f(x)|^p)^{1/p} 1^{1/q}$ and apply Hölder's ineq. $\endgroup$ Jul 23 '14 at 9:01
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Now I've thought that since $L^1 \subset L^p$ this is true also for f $\in L^p$ (can I do it? )

No because we don' t have the inclusion $\mathbb L^1(\mathbb R)\subset\mathbb L^p(\mathbb R)$ (consider $x^{1/p}\chi_{(0,1)})$.

However, we can use Jensen's inequality for the probability measure $\mu$ on the real line, given by $$\mu(B)=\frac 1{\lVert g\rVert_1}\int_B|g(x)|\mathrm dx.$$ We have $$\left(\int_{\mathbb R}|f(x-y)|\cdot |g(y)|\mathrm dy\right)^p =\left(\int_{\mathbb R}|f(x-y)|\mathrm d\mu(y)\right)^p\lVert g\rVert_1^p\leqslant\int_{\mathbb R}|f(x-y)|^p|\mathrm d\mu(y)\lVert g\rVert_1^p.$$ Now integrate over $x$ with respect to Lebesgue measure and use Fubini-Tonelli's theorem to conclude.

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