Problem related to $L^p$ space

Question

Problem Let $k: \mathbb R^{d\times d} \to \mathbb R^d$ be a measurable function such that there is $c>0$ with $$\sup_{x \in \mathbb R^d}\int |k(x,y)|dy \leq c, \space \sup_{y \in \mathbb R^d}\int |k(x,y)|dx \leq c$$

Show that for $1<p<\infty$, the function $K:L^p(\mathbb R^d) \to L^p(\mathbb R^d)$ given by $$K(f)(x)=\int k(x,y)f(y)dy$$

is well defined and uniformly continuous.

I got stuck trying to show both properties. First I would like to prove that $K(f) \in L^p(\mathbb R^d)$ for each $f \in L^p(\mathbb R^d)$, so take a function $f$ from that space and $$\int |K(f)(x)|^pdx=\int |\int (k(x,y)fy)dy|^pdx$$

I thought of applying Hölder's inequality to get $$\int |\int (k(x,y)fy)dy|^pdx \leq \int(\int |k(x,y)|^qdy)^{\frac{p}{q}})(\int|f(y)|^pdy)dx$$$$=\int|f(y)|^pdy(\int(\int |k(x,y)|^qdy)^{\frac{p}{q}}dx)$$

I don't know where to go from there.

As for uniform continuity I am lost. I would appreciate hints and suggestions to prove the two properties of the exercise. Thanks in advance.

Answer 1

Use (Riesz-Thorin) interpolation. Show that $\|K\|_{L^1(\Bbb R^d) \to L^1(\Bbb R^d)} \le c$ using $\sup_y \int |k(x,y)|\, dx \le c$, and show that $\|K\|_{L^\infty(\Bbb R^d)\to L^\infty(\Bbb R^d)} \le c$ using $\sup_x \int |k(x,y)|\, dy \le c$. These imply $\|K\|_{L^p(\Bbb R^d) \to L^p(\Bbb R^d)} \le c$ for all $1 < p < \infty$ by interpolation. This also gives well-definedness. For all $f,g\in L^p(\Bbb R^d)$, $$\|K(f) - K(g)\|_{L^p(\Bbb R^d)} = \|K(f-g)\|_{L^p(\Bbb R^d)} \le c\|f - g\|_{L^p(\Bbb R^d)}$$ So $K$ is uniformly continuous.

Answer 2

I'll write an alternative answer to kobe's, this solution relies basically on Hölder's inequality and Fubini-Tonelli:

First let's prove that the operator $K$ takes function from $L^p(\mathbb R^d)$ to functions on $L^p(\mathbb R^d)$:

Let $q$ be the conjugate of $p$, i.e., $\frac{1}{p}+\frac{1}{q}=1$. Then, $$||K(f)||_p=(\int_{\mathbb R^d}|\int_{\mathbb R^d}k(x,y)f(y)dy|^pdx)^{\frac{1}{p}}$$$$=(\int_{\mathbb R^d}|\int_{\mathbb R^d}(k(x,y))^{\frac{1}{p}}(k(x,y))^{\frac{1}{q}}f(y)dy|^pdx)^{\frac{1}{p}}$$ $$\leq (\int_{\mathbb R^d}\int_{\mathbb R^d}|(k(x,y))^{\frac{1}{p}}(k(x,y))^{\frac{1}{q}}f(y)|dy^pdx)^{\frac{1}{p}}$$

Now using Hölder's inequality conveniently, we have that the last expression is less than or equal to $$(\int_{\mathbb R^d} (\int_{\mathbb R^d}|k(x,y)||f(y)|^pdy)(\int_{\mathbb R^d}|k(x,y)|dy)^{\frac{p}{q}}dx)^{\frac{1}{p}}$$

With the hypothesis of the problem, this last integral is less than or equal to $$c^{\frac{1}{q}}(\int_{\mathbb R^d}\int_{\mathbb R^d}|k(x,y)||f(y)|^pdydx)^{\frac{1}{p}}$$

We can apply Tonelli to change the order of integration, so we get $$c^{\frac{1}{q}}(\int_{\mathbb R^d}\int_{\mathbb R^d}|k(x,y)||f(y)|^pdxdy)^{\frac{1}{p}}$$$$=c^{\frac{1}{q}}(\int_{\mathbb R^d}|f(y)|^p(\int_{\mathbb R^d}|k(x,y)|dx)dy)^{\frac{1}{p}}$$$$\leq c(\int_{\mathbb R^d}|f(y)|^pdy)^{\frac{1}{p}}< \infty$$

So we proved that $K(f) \in L^p(\mathbb R^d)$ and, moreover, that $$||K(f)||_p \leq c ||f||_p$$

To prove uniform continuity notice that $||K(f)-K(g)||_p=||K(f-g)||_p \leq ||f-g||_p$.