$K$ integrable on $\mathbb{R}$ implies $K$ bounded?

Question

Suppose $K:\mathbb{R} \rightarrow \mathbb{R}$ is a nonnegative function satisfying $\int_\mathbb{R}K(x)dx=1$ and $\int_\mathbb{R}K^2(x)dx<\infty$.

I know that if $K:\mathbb{R} \rightarrow \mathbb{R}$ is Riemann-integrable on $[a,b]$, then $K$ is a bounded function on $[a,b]$. But when the interval $[a,b]$ is the entire $\mathbb{R}$, it involves Lebesgue integration, and Lebesgue integrability does not necessarily implies boundness.

From the assumptions above, can I conclude $K$ is a bounded function on $\mathbb{R}$? Can you show me how could I argue that?

Thanks in advance.

Answer 1

No, you cannot. Define$$K(x)=\begin{cases}1&\text{ if }x\in[0,1]\\x&\text{ if }x=n\text{ for some natural }n\\0&\text{ otherwise.}\end{cases}$$Then your conditions hold, but $K$ is unbounded.

Answer 2

Jose Carlos Santos gave a good example, however his function is $L^{\infty}$ (bounded almost everywhere). For a non $L^{\infty}$ example, take: $$ f(x) = \begin{cases} \frac{1}{x^{1/4}} & x \in (0,1] \\ 0 & \text{otherwise} \end{cases} $$ Then, for any $M \in \mathbb{R}_{>0}$, we may find a set of nonzero measure $S$ such that for all $x \in S$ ,$f(x) > M$. However, the measure of the set on which this happens does tend to $0$.