what functions or classes of functions are $L^1$ but not $L^2$ We know that if a real function is in $L^2$ then it is in $L^1$, but the reverse is not necessarily true. So what are the examples of functions that are $L^1$ but not $L^2$, especially those intrinsically unbounded ones if any? Welcome more examples, or classes of examples. Thanks.
Neither containment holds in general.  On $\mathbb{R}$, the function $f$ such that $f(x)=1/x$ if $x\geq1$ and $f(x)=0$ otherwise is in $L^2\setminus L^1$, and the function $g$ such that $g(x)=1/\sqrt{x}$ if $0\lt x\leq1$ and $g(x)=0$ otherwise is in $L^1\setminus L^2$.

For bounded domains, $L^2\subset L^1$, because what keeps a function from being integrable on a bounded set is being too large, and squares make large numbers larger.  If $f$ is a bounded function in $L^1$, then $f$ is also in $L^2$, because what keeps a bounded function from being integrable is not going to zero fast enough, and squares make numbers go to zero faster.
Coming across these answers, it might be valuable to add the following explicit calculation:
Let $f \in L^1(\mathbb{R}^d) \cap L^{\infty}(\mathbb{R}^d)$. Then $f \in L^2(\mathbb{R}^d)$.
Indeed, denoting by $dx$ the Lebesgue measure on $\mathbb{R}^d$, we have
$$\int_{\mathbb{R}^d}|f|^2 = \int_{|f| \geq 1} |f|^2 + \int_{|f| &lt; 1} |f|^2 \leq \int_{|f| \geq 1} C^2 + \int_{|f| &lt; 1}|f| \leq C^2 \cdot dx(|f| \geq 1) + ||f||_{L^1} &lt; +\infty.$$ The first inequality uses the essential boundedness of $f$ and the last inequality uses $f \in L^1(\mathbb{R}^d),$ which yields $dx(|f| \geq 1) &lt; +\infty$.
On an unbounded domain $D$ (such as $D= \mathbb{R}^d$), the &quot;converse&quot;, i.e. $L^{\infty}(D) \cap L^2(D) \subseteq L^1(D)$ does not hold: While we can estimate
$$\int_D|f| \leq \int_{D \cap |f| &gt; 1} |f|^2+\int_{D \cap |f| \leq 1} |f| \leq ||f||_{L^2(D)}+dx(D\cap \{|f| \leq 1\}),$$we cannot conclude $f \in L^1(D)$, because the last summand is certainly not finite. As nicely explained in the above answers, this may be understood as insufficient fast decay to $0$ of $|f|$ on $D \cap \{|f| \leq 1\}$, while the decay of $|f|^2$ on the same set is fast enough.