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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?

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  • $\begingroup$ What is the natural metric? I feel like the one you wrote for $Y$ is the one I would have called 'natural.' Is it the one induced by the sup norm? $\endgroup$ – Alfred Yerger Dec 24 '17 at 22:50
  • $\begingroup$ @AlfredYerger I edited my question $\endgroup$ – Pumpkin Dec 24 '17 at 22:53
  • $\begingroup$ I edited your question so that the definition of the map $T$ is exposed to the title. Hope this does not harm your intention. $\endgroup$ – Sangchul Lee Dec 24 '17 at 23:21
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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.

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  • $\begingroup$ The metric was also wrong. I'm supposed to integrate those output functions over $[0,1]$, but this doesn't change the essence of the answer. I hope everything is OK now. $\endgroup$ – Alfred Yerger Dec 24 '17 at 23:11
  • $\begingroup$ I rephrased my answer in the hopes that it would be a little clearer this way. Essentially the idea is that I spotted the metric on the original functions inside the expression, bounded that, and then just evaluated the rest. $\endgroup$ – Alfred Yerger Dec 24 '17 at 23:21
  • $\begingroup$ @DavidC.Ullrich, I think this should clear it all up now. :) $\endgroup$ – Alfred Yerger Dec 24 '17 at 23:24
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$T$ 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 $\|T\| = \frac43$.

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