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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

$T$ 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 $\|T\| = \frac43$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.