Let $f \in \mathcal{L}^2(\mathbb{R})$ and $k \in \mathcal{L}^2(\mathbb{R \times\mathbb{R}})$. Consider the function


Prove that $g \in \mathcal{L}^2(\mathbb{R})$.

Here's my incomplete attempt: For $g$ to be in $\mathcal{L}^2(\mathbb{R})$, it needs to be measurable and $\int_\mathbb{R} g\,\text{d}\lambda<\infty$.


by some elementary known inequality I can't remember the name of. Then we can apply Hölder's inequality to $p=q=2$ to find that the expression in the RHS is $\leq$ than


What can we do after this? I feel like using Hölder's inequality is the way to go, but I'm not sure what's the next step.

  • $\begingroup$ the first sentence of your incomplete attempt is wrong $\endgroup$ – mathworker21 Jun 25 at 20:29
  • 2
    $\begingroup$ also, you applied Hölder with $p=q=2$ wrong. The square should have gone away. With the square gone, you're basically just done by Tonelli (change the second dummy variable from $y$ to $y'$ to avoid confusion). $\endgroup$ – mathworker21 Jun 25 at 20:30
  • $\begingroup$ also, "$\le$" reads as "less than or equal to", so you don't need to say "than" $\endgroup$ – mathworker21 Jun 25 at 20:32
  • $\begingroup$ With "first sentence", you mean that $\int_\mathbb{R} g\,\text{d}\lambda <\infty$ should be $\int_\mathbb{R} |g|^2\,\text{d}\lambda <\infty$, right? That was an oversight. $\endgroup$ – AstlyDichrar Jun 25 at 20:38
  • $\begingroup$ yes,,,,,,,,,,,, $\endgroup$ – mathworker21 Jun 25 at 20:38

To show that $g \in L^2(\mathbb R)$ you need to show to things:

  1. $g$ is measurable.
  2. $\Vert g \Vert_{\mathrm L^2(\mathbb R)} < \infty$.

The first assertion is clear from basic measure theory, since it follows from the hypothesis that $(x, y) \mapsto k(x,y) f(y)$ is measurable on the product space $\mathbb R \times \mathbb R$ and thus the function defined to be its integral in one variable is measurable.

The second assertion can be seen as follows: Since $k \in L^2(\mathbb R \times \mathbb R)$ you know that $k(x, \cdot) \in L^2(\mathbb R)$ for almosty every $x \in \mathbb R$. Thus, the Cauchy-Schwarz inequality tells you that \begin{align*} \left|\int_\mathbb{R}k(x,y)f(y)\,\text{d}\lambda(y)\right|^2 &= \left\vert \langle k(x, \cdot), f \rangle_{\mathrm L^2(\mathbb R)} \right\vert^2 \\ &\leq \Vert k(x, \cdot) \Vert_{\mathrm L^2(\mathbb R)}^2 \cdot \Vert f \Vert_{\mathrm L^2(\mathbb R)}^2 = \left(\int_\mathbb{R}\vert k(x,y) \vert ^2\,\text{d}\lambda(y) \right) \cdot \left(\int_\mathbb{R}\vert f(y) \vert ^2\,\text{d}\lambda(y) \right) \end{align*} for almost every $x \in \mathbb R$. Thus, you obtain \begin{align*} \Vert g \Vert_{\mathrm L^2(\mathbb R)}^2 &= \int_\mathbb{R}\left|\int_\mathbb{R}k(x,y)f(y)\,\text{d}\lambda(y)\right|^2\,\mathrm d\lambda(x) \\ &\leq \int_\mathbb{R} \left(\int_\mathbb{R}\vert k(x,y) \vert ^2\,\text{d}\lambda(y) \right) \cdot \left(\int_\mathbb{R}\vert f(y) \vert ^2\,\text{d}\lambda(y) \right) \,\mathrm d\lambda(x) \\ &= \int_{\mathbb{R} \times \mathbb{R}}\vert k(x,y) \vert ^2\,\text{d}\lambda^2(x, y) \, \cdot \, \int_\mathbb{R}\vert f(y) \vert ^2\,\text{d}\lambda(y) \\ &= \Vert k \Vert_{\mathrm L^2(\mathbb R \times \mathbb R )}^2 \cdot \Vert f \Vert_{\mathrm L^2(\mathbb R)}^2 < \infty. \end{align*} I hope the arguments got clear to you :)

  • $\begingroup$ This is what I got after correcting the mistake mathworker21 pointed out (Hölder's inequality with $p=q=2$ is the C-S inequality). Thanks! $\endgroup$ – AstlyDichrar Jun 25 at 23:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.