Please give me a hint not whole solution:
The Problem: Let $K(x,t):\mathbb{R}^{2}\to \mathbb{R}$ be a non-negative measurable function such that $$F(x) = \int\limits_{\mathbb{R}} K(x,t)dt \in L(\mathbb{R})$$
- (a) For $f \in L(\mathbb{R})$ show that $$g(x) = \int\limits_{\mathbb{R}}K(x,t)f(t)dt$$ is measurable and $g \in L(\mathbb{R}).$
What I have done:
Since $F(x)\in L^{1}(\mathbb{R})$ then
$$ \int\limits_{\mathbb{R}}F(x)dx<\infty \Rightarrow \int\limits_{\mathbb{R}}\vert \int K(x,t)dt \vert dx<\infty$$
but $K(x,t)\in \mathbb{R}^{+} $ so
$$ \Rightarrow \int\limits_{\mathbb{R}} \int K(x,t)dt dx<\infty\Rightarrow \int\limits_{\mathbb{R}} (\int K(x,t)dx) dt<\infty$$
Since $\int K(x,t)dx>0$ so we can conclude that $\int K(x,t)dx<\infty.$
Now we want to show to show that $g(x)\in L^{1}(\mathbb{R})$ so we know that $\vert K(x,t)f(t)\vert$ is measurable so I can use Fubini theorem so
$$ \int\limits_{\mathbb{R}}\vert \int K(x,t)f(t)dt \vert dx\leq \int\limits_{\mathbb{R}} \int K(x,t)\vert f(t)\vert dt dx \\= \int\int K(x,t)\vert f(t)\vert dx dt= \int\int K(x,t)\vert f(t)\vert dx dt\leq M\int K(x,t)dx $$ where $\int\vert K(x,y)\vert dt<M$. Therefore, $$g\in L^{1}(\mathbb{R}).$$
- (b) Compute $$\sup\limits_{\Vert f\Vert_{1}\leq 1} \int\limits_{\mathbb{R}^{2}} K(x,t) \vert f(t)\vert dx\, dt. $$
What I have done: $$\sup\limits_{\Vert f\Vert\leq 1} \int\limits_{\mathbb{R}^{2}} K(x,t) \vert f(t)\vert dx\, dt\leq \sup\limits_{\Vert f\Vert\leq 1} \int\limits_{\mathbb{R}}\vert f(t)\vert\int K(x,t) dx\, dt=\int K(x,t)dx< +\infty$$
- (c) Evaluate $$ \lim\limits_{x\to \infty} \int_{\mathbb{R}} K(x,t)f(t)dt$$ where $f \in L(\mathbb{R}).$
For this Part, I couldn't do anything.
(a) The conclusion here is false. Proof: Let $K(x,t) = t^{-1/2}\chi_{(0,1)^2}(x,t).$ Then $F(x) = 2\chi_{(0,1)}(x),$ so $F\in L^1.$ But if we let $f(t) = t^{-1/2}\chi_{(0,1)}(t),$ then $f\in L^1,$ but $g(x) = \infty$ for $x\in (0,1).$ Thus $g\notin L^1.$
(b) The example I gave in (a) shows that the supremum here can be $\infty.$ You made a mistake in the last step of "What I have done".
(c ) There are functions $g\in L^1(\mathbb R)$ such that $\lim_{x\to \infty} g(x)$ fails to exist; in fact the behavior of $g$ can be rather wild at $\infty.$ Fix such a $g.$ Then define
$$K(x,t) = \frac{g(x)}{1+t^2}, \,\, f(t) = \frac{1}{1+t^2}.$$
Then
$$\tag 1 \int_{\mathbb R} K(x,t)f(t)\,dt = g(x)\int_{\mathbb R} \frac{1}{(1+t^2)^2}\,dt.$$
Since $\lim_{x\to \infty} g(x)$ fails to exist, the limit in $(1)$ fails to exist.
One question for you: Are sure this problem is stated correctly? Where is it from? It doesn't feel right. At any rate, I've answered the questions as stated.