Continuous bounded functions in $L^1$ 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?
 A: 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.
