Is the mapping $T : x \mapsto \int_{0}^{\bullet} \tau^{-1/2}x(\tau)\,d\tau$ uniformly continuous? Consider $X = C([0,1])$ with its natural metric induced by  $\| \cdot \|_{\sup}$ and $Y = C([0,1])$ with the metric $d_1(x,y) = \int^1_0 |x(t)-y(t)| \, dt$. Let
$$T: X\to Y : x(t) \mapsto y(t) = \int_0^t \frac{1}{\sqrt \tau} \ x(\tau) \, d\tau$$
Is the mapping T uniformly continuous?
Definition. $T:(X,d_x) \to (Y,d_y)$ is uniformly continuous if $\forall \epsilon$, $\exists \delta = \delta(\epsilon)$ :
$$ \quad T(B(a,\delta)) \subset B(T_a,\epsilon), \qquad \forall a \in X$$
Usage of $\delta$ in the definition confuses me how can I prove this?
 A: I believe this is uniformly continuous. We can estimate 
$$d(T(x(t)), \  T(y(t))) = \int_0^1\bigg|\int_0^t \frac{1}{\sqrt \tau}\big[x(\tau) - y(\tau) \big] d\tau  \bigg | dt \leq \int_0^1\int_0^t \bigg|\frac{1}{\sqrt \tau} \bigg| \  \delta \ d\tau  \ dt$$
Where we assume that the distance between $x(t)$ and $y(t)$ is $\delta$. We will determine exactly what $\delta$ should be in terms of $\epsilon$, and this will complete our proof. Since we're working in the sup norm in $X$. Now it just remains to figure out how to choose $\delta$ to make this quantity smaller than $\epsilon$. Since $\frac{1}{\sqrt\tau}$ is positive on $(0,1)$, we drop the bars.
$$d(T(x(t)), \ T(y(t))) \leq \delta \int_0^1\int_0^t \frac{1}{\sqrt \tau} d \tau \ dt$$
And this is easily evalauted to be $4\delta/3$. So taking $\delta = 3\epsilon/4$, we're done. 
Since $x(t)$ and $y(t)$ were arbitrary, this proves uniform continuity.
A: $T$ is indeed uniformly continuous.
Notice that the function $T$ is in fact a linear map between two normed spaces:
$$T : \left(C[0,1], \|\cdot\|_\infty\right) \to \left(C[0,1], \|\cdot\|_1\right)$$
where $\|f\|_1 = \int_0^1 \left|f(x)\right|\,dx$.
Recall that a linear map is uniformly continuous if and only if it is bounded. Therefore, it only remains to show that $T$ is bounded:
\begin{align}
\|Tf\|_1 &= \int_0^1\int_0^t \frac{\left|f(x)\right|}{\sqrt{x}}\,dx\,dt\\
&= \int_0^1\int_x^1 \frac{\left|f(x)\right|}{\sqrt{x}}\,dt\,dx\\
&= \int_0^1\left(\int_x^1 dt\right)\frac{\left|f(x)\right|}{\sqrt{x}}\,dx\\
&= \int_0^1\frac{1-x}{\sqrt{x}}\left|f(x)\right|\,dx\\
&\le \int_0^1\frac{1-x}{\sqrt{x}}\left\|f\right\|_\infty\,dx\\
&= \|f\|_\infty\int_0^1\frac{1-x}{\sqrt{x}}\,dx\\
&= \frac43 \|f\|_\infty
\end{align}
Therefore, $T$ is bounded, and hence uniformly continuous.
Furthermore, $\|T\| \le \frac43$ and for $f \equiv 1$ we have $\|Tf\|_1 = \frac43$ so $\|T\| = \frac43$.
