Define $T:X \to X$ by $ Tf(x) = x \int_0^x f(t) dt $ Prove that the inverse $T^{-1}:ran(T) \to X$ exists but is not bounded. The question is the following

Let $X = (C([0,1]),||\cdot||_\infty)$. Define $T:X \to X$ by
  $$
Tf(x) = x \int_0^x f(t) dt
$$
  Prove that the inverse $T^{-1}:ran(T) \to X$ exists but is not bounded.

My attempt is the following:
Now, define
$$
F(x) = \int_0^x f(t) dt
$$
then we have $F' = f$ and $g \in Tf$ if and only if 
$$
g(x) = xF(x) \quad \ \implies \quad \frac{g(x)}{x} = F(x)
$$
Take the derivative. Since the $g(x)$ is the product of two differentiable functions which is differentiable, we have
$$
\frac{xg'(x)-g(x)}{x^2} = f(x)
$$
and this defines the inverse $T^{-1}$. 
My question is that whether my attempt is correct? If so, how to show that it is not bounded?
 A: Existence of $T^{-1}$: for linear operator $T$ to be invertible all we need to prove is that $\mathrm{Ker} \,T=\{0\}$. Let $f\in \mathrm{Ker}\,T$, then $Tf=0$, i.e. $x\int\limits_0^xf(t)\,\mathrm{d}t=0$ for all $x\in [0,1].$ Then
$\int\limits_0^xf(t)\,\mathrm{d}t=0$ for all $x\in (0,1]$ and by taking the derivative in both sides of the above equation:
$$\left(\int\limits_0^xf(t)\,\mathrm{d}t\right)'=0\iff f(x)=0$$
for all $x\in (0,1]$, so $f\equiv 0$ by continuity of $f.$
Formula for $T^{-1}:$ your work is correct for $x\in (0,1]$, so:
$$T^{-1}(g)(x)=\frac{g'(x)}{x}-\frac{g(x)}{x^2},~~x\in (0,1].$$
$T^{-1}$ is unbounded: we need a sequence $(g_n)$ in $\mathrm{Im}\,T$ such that $\|g_n\|_\infty\leqslant 1$ for all $n$, but $\|T^{-1}g_n\|_\infty \to +\infty$.
Take $g_n:[0,1]\to \mathbb{R}:g_n(x)=x^{n+1}$, so every member of $(g_n)$ belongs in $\mathrm{Im}\,T$, we have $\|g_n\|_\infty=1$ and for $x\in (0,1]:$
$$T^{-1}(g_n)(x)=\frac{g_n'(x)}{x}-\frac{g_n(x)}{x^2}=(n+1)x^{n-1}-x^{n-1}=nx^{n-1},$$
so $$\|T^{-1}g_n\|_\infty\geqslant \sup_{x\in (0,1]}nx^{n-1}=n.$$
A: To prove the boundedness, consider $f_{n}=n^{1/2}$ for $0\leq x\leq 1/n$ and $f_{n}=1/x^{1/2}$ for $1/n\leq x\leq 1$, then $Tf_{n}(x)\leq\displaystyle\int_{0}^{1}f_{n}(t)dt=2-1/n^{1/2}<2$, and hence $\|Tf_{n}\|_{L^{\infty}[0,1]}\leq 2$ but $\|f_{n}\|_{L^{\infty}[0,1]}=n^{1/2}\rightarrow\infty$.
