# Compact integral operator on $C([0,1])$

On space $X=C([0,1])$ we are given an operator $T:X \to X$ with $$Tf(x)=\int_0^x{f(y)}dy$$ I proved using Arzela-Ascoli that the operator is compact.

The other question is to prove that $T(B_X)$ is not closed (although its closure is compact) where $B_X$ is closed unit ball in $X$. I was trying to construct a sequence $(f_n) \in B_X$ where $Tf_n \to f$ and $f$ is not in $T(B_X)$ but I failed. I tried with some polynomials, trigonometric functions, exponential, but no success. Any idea how to construct the sequence or on some other way prove the above fact.

• Hint: If $f\in C([0,1])$, then $Tf(0)=0$ and $Tf$ is $C^1$. So try to find a sequence of $C^1$ functions $\{g_n\}$ such that $g_n(0)=0$ and $g_n$ converges to a non-$C^1$ function $g$. – Joey Zou Nov 1 '16 at 9:37
• These functions $g_n$ should be image of functions $f_n \in B_X$, but $g$ is not image of any such function since it is not $C^1$? – Marko Rajkovic Nov 1 '16 at 9:52
• I guess the functions $g_n(x)=\sqrt{(x-1/2)^2 + 1/n}-\sqrt{1/4 +1/n}$ are ok. – Marko Rajkovic Nov 1 '16 at 10:42