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Let $T(f)(x):=\int_{\mathbb{R}^d} k(x,y)f(y) dy$ where $k_1(x)= \int_{\mathbb{R}^d} |k(x,y)|dy$ and $k_2(y)= \int_{\mathbb{R}^d} |k(x,y)|dx$ are two $L^{\infty}$ functions. Then I want to show that $T: L^2 \rightarrow L^2$ has $||T||\le \sqrt{||k_1||_{L^{\infty}}||k_2||_{L^{\infty}}}.$

I was able to show this inequality without the square-root, i.e.

$$||T||_{L^2} = \sup_{||\phi||_2=||\psi||_2=1} \langle T \psi, \phi \rangle$$ $$\le \sup_{||\phi||_2=||\psi||_2=1} \int_{\mathbb{R}^{2d}} |k(x,y)| |\phi(x)| |\psi(y)| dx dy$$ and now by Cauchy schwartz $$\le \sup_{||\phi||_2=||\psi||_2=1} \int_{\mathbb{R}^{2d}} |k(x,y)| |\phi(x)|^2 dx dy\int_{\mathbb{R}^{2d}}|k(x,y)||\psi(y)| dx dy$$ $$\le \sup_{||\phi||_2=||\psi||_2=1} \int_{\mathbb{R}^{d}}|k_1(y)| |\phi(y)|^2 dy \int_{\mathbb{R}^{d}}|k_2(x)| |\phi(x)|^2 dx$$

$$\le ||k_1||_{\infty} ||k_2||_{\infty}$$

Now, I was wondering whether the more inequality stated on the exercise sheet is still correct or whether there has to be some error?

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You forgot a square root in the Cauchy-Schwarz inequality. To be more precise: Choose $\phi, \psi \in L^{2}(\mathbb{R}^d)$ with $\lvert \lvert \phi \rvert \rvert_2 = \lvert \lvert \psi \rvert \rvert_2 = 1$. Then \begin{align} \lvert \langle T \psi, \phi \rangle \rvert = \Bigg| \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} k(x,y) \psi(x) \phi(y) \, \textrm{d}y \, \textrm{d}x \Bigg| \leq \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \big( \lvert k(x,y) \rvert^{1/2} \lvert \phi(x) \rvert \big) \big( \lvert k(x,y) \rvert^{1/2} \lvert \psi(y) \rvert \big) \, \textrm{d}y \, \textrm{d}x. \end{align} Now, we can apply Cauchy-Schwarz' inequality (or Hölder inequality) and Tonelli's theorem, and we obtain that \begin{align} \lvert \langle T \psi, \phi \rangle \rvert &\leq \Bigg( \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \lvert k(x,y) \rvert \lvert \phi(x) \rvert^{2} \, \textrm{d}y \, \textrm{d}x \Bigg)^{1/2}\Bigg( \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \lvert k(x,y) \rvert \lvert \psi(y) \rvert^{2} \, \textrm{d}x \, \textrm{d}y \Bigg)^{1/2}\\ & = \Bigg( \int_{\mathbb{R}^d} k_1(x) \lvert \phi(x) \rvert^{2} \, \textrm{d}x \Bigg)^{1/2}\Bigg( \int_{\mathbb{R}^d} k_2(y) \lvert \psi(y) \rvert^{2} \, \textrm{d}x \Bigg)^{1/2}\\ & \leq \Big( \lvert \lvert k_1 \rvert \rvert_{L^{\infty}(\mathbb{R}^d)} \lvert \lvert k_2 \rvert \rvert_{L^{\infty}(\mathbb{R}^d) } \lvert \lvert \phi \rvert \rvert_2 \lvert \lvert \psi \rvert \rvert_2 \Big)^{1/2} = \sqrt{ \lvert \lvert k_1 \rvert \rvert_{L^{\infty}(\mathbb{R}^d)} \lvert \lvert k_2 \rvert \rvert_{L^{\infty}(\mathbb{R}^d) } }. \end{align} As this is true for all $\phi, \psi \in L^{2}(\mathbb{R}^d)$ with $\lvert \lvert \phi \rvert \rvert_2 = \lvert \lvert \psi \rvert \rvert_2 = 1$ the claim follows.

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