A function which is in $L^1$ but does not belong to $L^\infty$

Question

Can someone give me an example of an $L ^1$ function which does not belong to $L^\infty$. In fact we look at $L^1(\Omega,\mathcal{F},P)$, where $(\Omega,\mathcal{F},P)$ denotes a probability space. Of course the function should be unbounded but the integral should exist. Clearly, we can embed $L^\infty$ into $L^1$ in this case. Thank you.

Answer 1

A simple example is given by $$f(x)={1\over \sqrt{x}}=x^{-1/2},\qquad 0<x\leq 1$$ and $f(x)=0$ otherwise.

Or you can use $g(x)=\log(x)$ for $0<x\leq 1$, with $g(x)=0$ otherwise.

Answer 2

Explicit example can be given when we specify the measure space, like $\frac{\log |x|}{1+x²}$ on the real line.

I will give a characterisation when $L^1\subset L^{\infty}$:

Let $(X,\mathcal F,\mu)$ a measure space. We have $L^1(\mu)\subset L^{\infty}(\mu)$ if and only if we can find a positive constant $c$ such that for $A\in\mathcal F$, either $ \mu(A)=0$ or $\mu(A)\geq c$.

If we have $L^1(\mu)\subset L^{\infty}(\mu)$, then the inclusion is linear and continuous (we can see this by the closed graph theorem), so we can find $K>0$ ($K=0$ in trivial cases) such that $\lVert f\rVert_{\infty}\leq K\cdot\lVert f\rVert_1$ for each $f\in L^1$. Now if $A$ is of finite measure, either its measure is $0$ or $\geq \frac 1K$.

Conversely, if $f\in L^1$, we can choose $\alpha$ such that $\mu\{|f|\geq \alpha\}<c$, hence $|f|\leq \alpha$ almost everywhere.

Answer 3

An example for the space $X= (0,1]$ is $\dfrac{1}{\sqrt{x}},$ which has infinite $L^{\infty}$ norm because as $x\to 0^+$ the function $1/\sqrt{x}\to \infty,$ but has $L^1$ norm of $\displaystyle \int^1_0 \frac{1}{\sqrt{x}} dx = 2.$

If you want an example for $X=\mathbb{R}$ then $1_{(0,1]} \dfrac{1}{\sqrt{x}}$ works similarly.