Then $K$ is a) compact b) not compact?

Question

Let $K \subset C[0, 1]$ (with the usual sup-norm metric) be defined by $$K = \bigg\{f \in C[0, 1]\mid \int_0^1 f(t) \, dt = 1 \text{ and } f(x) \geq 0 \text{ for all } x \in [0, 1]\bigg\}$$ Then $K$ is

a) compact

b) not compact

i thinks this will compact because K is closed and bounded by hein Borel thorem it is compact

is My answer is correct or not ?

Answer 1

In fact, $K$ fails to be bounded and therefore fails to be compact. Note for instance that the family of functions defined by $$ f_n(x) = \begin{cases} n & x \leq \frac 1n - \epsilon_n\\ \text{[linear function]} & \frac 1n - \epsilon_n < x < \frac 1n + \epsilon_n\\ 0 & x \geq \frac 1n + \epsilon_n \end{cases} $$ (where for each $n$, $\epsilon_n$ is chosen to be less than $1/n$) is a subset of $K$ which is unbounded in sup-norm.

Answer 2

$K$ is not even bounded you may take another example say,

$f_n(x)=nx^{n-1} \forall n>1$. See $\|f_n\|_{\infty}=n \to \infty$