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.

  • $\begingroup$ Are you asking if the set $F$ is bounded with respect to the supremum norm? $\endgroup$ Jan 22, 2021 at 16:12
  • $\begingroup$ @CameronWilliams, yes, under the sup norm. $\endgroup$ Jan 22, 2021 at 16:13
  • $\begingroup$ You know that $$f_n(x) = x^n$$ is not equicontinuous. How would you modify this? $\endgroup$
    – macton
    Jan 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 $1/n$. $\endgroup$ Jan 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? $\endgroup$ Jan 22, 2021 at 17:50

1 Answer 1


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.


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