Boundedness in $(M_{\phi}f)(x)=\phi(x)f(x)$ and $(Kf)(x)=\int_a^bk(x,t)f(t) dt$ I'm currently practicing (for a comprehensive exit exam) how to prove something is bounded. Here are 2 questions I'm concerned about:


*

*Let $\phi$ be a continuous function on the interval $[a,b]$.
Define $M_{\phi}:L^2[a,b] \to L^2[a,b]$ by $(M_{\phi}f)(x)=\phi(x)f(x)$
Show that $M_{\phi}$ is a bounded linear operator on $L^2[a,b]$



*

*Let $k(x,t)$ be a continuous function on the square $[a,b]$ x $[a,b]$
Define $K:L^2[a,b] \to L^2[a,b]$ by $(Kf)(x)=\int_a^bk(x,t)f(t) dt$ 
Show that $K$ is a bounded linear operator on $L^2[a,b]$

My Approach: 


*

*$\left |(M_{\phi}f)(x) \right |=\left |\phi(x)f(x) \right |$. Since $\phi$ is a continuous function on the interval $[a,b]$, then $|\phi(x)| \le k_0$ where $k_0$ is a real number for all $x$ in the space. Therefore $$\left |(M_{\phi}f)(x) \right |=\left |\phi(x)f(x) \right | \le k_0 ||f||$$

*Similarly, $|k(x,t)| \le k_0$ for all $(x,t) \in J=[a,b]*[a,b]$. Therefore, $$\left| (Kf)(x)\right |=\left|\int_a^bk(x,t)f(t)dt \right| \le k_0||f||$$


Is this the proper way to prove these? Thank you.
 A: Not quite, sorry. To show an operator $T$ on $L^2$ to itself is bounded, you must show that there is a constant $C$ such that
$$\|Tf\|_{L^2} \le C \|f\|_{L^2}$$
for all $f \in L^2$. So pointwise bounds aren't quite what you're looking for.

But your proof of (1) can be adapted: Choosing $k_0$ large enough that $|\varphi(x)| \le k_0$ for all $x$, we see that
$$|M_{\varphi} f|(x) = |\varphi(x)| |f(x)| \le k_0 |f(x)|$$
Now square this inequality and integrate to find that
$$\int_a^b |M_{\varphi}f|^2 dx \le k_0^2 \int_a^b |f|^2 dx$$
Taking a square root shows that $\|M_{\varphi} f\|_{L^2} \le k_0 \|f\|_{L^2}$.
A: There are several approaches that will work for $K$. I'll choose the direct approach. Start with the Cauchy-Schwarz inequality
$$
            |Kf(x)|^{2} \le \left[\int_{a}^{b}|k(x,t)||f(t)|\,dt\right]^{2}
           \le \int_{a}^{b}|k(x,t)|^{2}\,dt\int_{a}^{b}|f(t)|^{2}\,dt=\int_{a}^{b}|k(x,t)|^{2}\,dt\cdot\|f\|^{2}_{L^{2}}.
$$
Integrating with respect to $x$ gives
$$
    \|Kf\|_{L^{2}}^{2} \le \int_{a}^{b}\int_{a}^{b}|k(x,t)|^{2}\,dx\,dt\cdot\|f\|^{2}_{L^{2}}.
$$
Because $k$ is continuous on $[a,b]\times[a,b]$, then it is bounded by some constant $M$ on that region. So, $\|Kf\| \le M(b-a)\|f\|$. Notice that $K$ is bounded under the weaker assumption that $k \in L^{2}([a,b]\times[a,b])$.
