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

$$g(x)=\int_\mathbb{R}k(x,y)f(y)\,\text{d}\lambda(y)$$

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 $\int_\mathbb{R} g\,\text{d}\lambda<\infty$.

$$\int_\mathbb{R}\left|\int_\mathbb{R}k(x,y)f(y)\,\text{d}\lambda(y)\right|^2\,\text{d}\lambda(x)\leq\int_\mathbb{R}\left(\int_\mathbb{R}\left|k(x,y)f(y)\right|\,\text{d}\lambda(y)\right)^2\,\text{d}\lambda(x)$$

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

$$\int_\mathbb{R}\left(\int_\mathbb{R}\left|k(x,y)\right|^2\,\text{d}\lambda(y)\int_\mathbb{R}\left|f(y)\right|^2\,\text{d}\lambda(y)\right)^2\,\text{d}\lambda(x)$$

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.

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  • $\begingroup$ the first sentence of your incomplete attempt is wrong $\endgroup$ – mathworker21 Jun 25 at 20:29
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    $\begingroup$ 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
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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 :)

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  • $\begingroup$ 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

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