6.V The space $L_{\infty}(X,X,\mu)$ is contained in $L_{1}(X,X,\mu)$ if and only if $\mu(X)<\infty$. If $\mu(X)=1$ and $f\in L_{\infty}$ ,then $$||f||_{\infty}=\lim_{p\to \infty}{||f||_{p}}$$
For the first part, if containment is true then $$\because f=1$$ a constant function, is contained in $L_{\infty}$. From our assumption it will also be contained in $L_{1}$. $$\therefore \int|f|d\mu<\infty$$ $$\int d\mu <\infty$$ $$\mu(X)<\infty$$
Conversly, if $\mu (X)<\infty$. Let $f\in L_{\infty}$ be an arbitrary element,$$\therefore ||f||_{\infty}<\infty$$ Also, there exists a measurable set $E$ with $\mu(E^{\complement}) =0$ such that for every $$\forall x\in E$$ $$|f(x)|\leq||f||_{\infty}$$ Consider, $$\int|f|d\mu = \int_{E}|f|d\mu + \int_{E^{\complement}}|f|d\mu$$ $$\int|f|d\mu = \int_{E}|f|d\mu + 0$$ $$\int|f|d\mu < \int_{E}||f||_{\infty}d\mu$$ $$\int|f|d\mu < ||f||_{\infty}\int_{E}d\mu$$ $$\int|f|d\mu < ||f||_{\infty}\mu(E)$$ $$\because \mu(E)<\mu(X)<\infty$$ $$\int|f|d\mu < \infty $$ $$\therefore f \in L_{1}$$ Hence,$$L_{\infty}\subset L_{1}$$ For the second part of the question, I was able to show that $$||f||_{p}<||f||_{\infty},$$ using similar steps as used to prove that $||f||_{1} < \infty$, for first part of the question. I think that it might come handy for solving the second part.
But i do not know, how to show that it is a convergent sequence and it converges to $||f||_{\infty}$.
