Is a continuous function in $L^1$bounded? I know that continouos functions are always bounded on a compact intervall. But how do I prove it?


If we study $L^1(0,\infty)$ or $L^1(\Bbb R)$, then you can have an unbounded continuous integrable function.

We can build it in the following way: let's start with $f= 0$. Then we add positive continuous "bumps" at $n=1,2$,etc with each "bump" being higher, say, of magnitude $n$, but adding only $1/n^2$ to the integral. Like this our function remains integrable, continuous, yet unbounded.

One can explicitly build such a function, I'll outline the important details

First, we take $$g(x)=\begin{cases} e^{-\frac{1}{1-x^2}},&|x|<1,\\0,&|x|\ge 1.\end{cases}$$ It's possible to show that this function is $\mathcal C^{\infty}(\Bbb R)$, its support is $[-1,1]$, it's positive, and its integral is finite (let's call it $I$). Its supremum is $e^{-1}$.

Now let's study $$g_n(x):=ng\left( (x-n)n^3 \right).$$ It's still continuous, positive, its support is $[n-n^{-3}, n+n^{-3}]$. Its integral is $\frac{I}{n^2}$, and its supremum is $ne^{-1}$.

We take the sum $$G(x):=\sum_{k\ge 3}g_k(x).$$ It's possible to show that $G$ is continuous, unbounded, positive, integrable.

  • $\begingroup$ I see. Thats pretty nice, actually. $\endgroup$ – Vishal Gupta Aug 18 '13 at 9:19

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