Integral Operator: Linear and Continuous

Question

I'm studying the following proposition:

An integral operator $Tf(y):= \int_Ak(x,y)f(x)dx$ is linear, and is continuous on the the following spaces:

$$ \sup_{x \in A, \>y\in B} |k(x,y)| < \infty \implies T:L^1(A) \rightarrow L^{\infty}(B) \tag1$$

$$ \int_B \sup_{x \in A} |k(x,y)|dy \lt \infty \implies T:L^1(A) \rightarrow L^1(B) \tag2 $$

Q1) In the proposition above when it says that: an integral operator $T$ is linear and, is continuous on the following spaces - what space is it referring to? normed space $L^\infty (B)$ in statement (1) ? normed space $L^1 (B)$ in statement (2)?

Q2) The implication arrows in statement (1) and (2) confuse me. What role does the left side of the implication arrow plays in the proposition? Would the proposition still be the same if the left side of the implication arrow is removed:

$$ T:L^1(A) \rightarrow L^{\infty}(B) \tag1$$

$$ T:L^1(A) \rightarrow L^1(B) \tag2 $$

After the proposition, the book that I used give an example: that Fourier Transform is an operator $F: L^1(\mathbb{R}) \rightarrow L^\infty (\mathbb{R})$

Given the above sentence, would it be correct to say that $Ff(y)$ is linear and continuous without given that e.g. (I guess:) $\sup_{x \in A, \>y\in B} |k(x,y)| < \infty$

Hope I'm being clear with my questions as I struggle to understand the proposition above. Thank you!

Answer 1

You have to show that if $\sup_{x \in A, y \in B}\vert k(x,y)< \infty$ the operator $T$ is a continous map from $L^1(A) \to L^\infty(B)$. More precisely you have to show the existence of a $C > 0$ such that for every $f \in L^1(A)$ the inequality $$ \sup_{y \in B} \vert Tf(y) \vert \leq C \Vert f \Vert_{L^1(A)}$$ holds. To find such a $C$ you need the left side of the implication arrow.

Answer 2

The correct way of stating the proposition is the following: 1) $T$ is a bounded operator from $L^{1}(A)$ to $L^{\infty}(B)$ if $sup_{x,y}|k(x,y)| <\infty$.

2). $T$ is a bounded operator from $L^{1}(A)$ to $L^{1}(B)$ if $\int_B \sup_x|k(x,y)|dy <\infty$.

The language used in the book is not a standard one and it has caused unnecessary confusion in your mind. Yes it is correct to say that FT is a bounded operator from $L^{1}(\mathbb R)$ to $L^{\infty}(\mathbb R)$.