Let $F= \{ f \in C([0,1]) : \int_0^1 |f(x)| dx \leq 1 \}$ then does $F$ bounded and equicontuous?
I am not sure how can I relate the function values with the integral of the function.
$\begingroup$
$\endgroup$
7
-
$\begingroup$ Are you asking if the set $F$ is bounded with respect to the supremum norm? $\endgroup$– Cameron WilliamsJan 22, 2021 at 16:12
-
$\begingroup$ @CameronWilliams, yes, under the sup norm. $\endgroup$– math_for_everJan 22, 2021 at 16:13
-
$\begingroup$ You know that $$f_n(x) = x^n$$ is not equicontinuous. How would you modify this? $\endgroup$– mactonJan 22, 2021 at 16:14
-
$\begingroup$ Can you construct a sequence of positive functions that have integral $1$ but have norms increasing to infinity? Hint: consider a triangle that has height $2n$ and width $1/n$. $\endgroup$– Cameron WilliamsJan 22, 2021 at 16:14
-
$\begingroup$ @CameronWilliams. I remember these functions, I constructed them their areas are 1, and they are not bounded, but are they not equicontinuous, correct? $\endgroup$– math_for_everJan 22, 2021 at 17:50
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
Consider $$f_n(x)=(n+1)x^n$$
Notice that for $x \neq 1$, $$ |f_n(1) - f_n(x)| = (n+1)(1 - x_0^n) $$ and as $x^n \to 0$ for $x \neq 1$, we see that the difference $|f_n(1) - f_n(x)|$ will not be bounded as $n$ came large, therefore this family is not equicontinuous.
