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Let $\mathcal{C}$ be the space of continuous functions on $[0,1]$ equipped with the norm $\|f\|=\int^{1}_{0}|f(t)|dt$. Define a linear map $\mathcal{T}:\mathcal{C}\rightarrow \mathcal{C}$ by $$ \mathcal{T}f(x)=\int^{x}_{0}f(t)dt. $$ Show that $\mathcal{T}$ is well-defined and bounded and determine the value of $\|\mathcal{T}\|_{\text{op}}$.


I proved the first two parts myself, but I am having trouble with determining the value of $\|\mathcal{T}\|_{\text{op}}$. I was able to show that it is bounded by $1$ though. Observe that

$$\|\mathcal{T}f\|=\int^{1}_{0}\left|\int^{x}_{0}f(t)dt\right|dx\leq \int^{1}_{0}\|f\|dx = \|f\| $$ Therefore,

$$ \|\mathcal{T}\|_{\text{op}} = \underset{\|f\|=1}{\sup}\frac{\|\mathcal{T}f\|}{\|f\|}\leq \underset{\|f\|=1}{\sup}\frac{\|f\|}{\|f\|}=1 $$

I tried seeing if I could then construct a function where the operator equals $1$, but I've had no success. Anybody have any solutions or hints? Any help is appreciated.

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    $\begingroup$ Sorry, had to make another edit due to a typo... $\endgroup$ – Joe Man Analysis Nov 14 '18 at 2:21
  • $\begingroup$ see volterra integral operator $\endgroup$ – qbert Nov 14 '18 at 3:08
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    $\begingroup$ Maybe, take $f_n=1_{[0,1/n]}$. Compute the norm of $f_n$ and of $Tf_n$. $\endgroup$ – Shalop Nov 14 '18 at 3:17
  • $\begingroup$ @Shalop. That's not a sequence of continuous functions. $\endgroup$ – md2perpe Nov 14 '18 at 9:32
  • $\begingroup$ @md2perpe fine. Then take $f_n(x)=\max\{1-nx,0\}$ I suppose. And $(1-x)^n$ probably works too. It’s a bit silly since continuous functions are dense in L^1 anyways, so it doesn’t matter much. $\endgroup$ – Shalop Nov 14 '18 at 12:43
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For $n\in \Bbb N:$ Let $K_n=\frac {1}{n+1}+\frac {1}{(n+1)^2}.$ Let $f_n(x)=n+1$ for $x\in [0,\frac {1}{n+1}].$ Let $f_n(x)=0$ for $x\in [K_n,1].$ Let $f_n(x)$ be linear for $x\in [\frac {1}{n+1},K_n].$

We have $\|f_n\|=1+\frac {1}{2(n+1)}.$

For $x\in [\frac {1}{n+1},1]$ we have $(Tf_n)(x)\geq (Tf_n)(\frac {1}{n+1})=1.$ $$\text {So }\quad \|Tf_n\|\geq \int_{1/(n+1)}^1 (Tf_n)(x)dx\geq \int_{1/(n+1)}^1 1\cdot dx=$$ $$=1-\frac {1}{n+1}.$$

$$\text {So} \quad \frac {\|Tf_n\|}{\|f_n\|}\geq \frac {1-\frac {1}{n+1}}{ 1+\frac {1}{2(n+1) }}.$$

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It might be hard (or impossible) to find a function for which $ ||\mathcal{T}f||$ is exactly equal to $||f||$, but you only need to find a sequence of functions $f_n$, such that $$\frac{||\mathcal{T}f_n||}{||f_n||} \to 1\quad \text{as } n \to \infty.$$

Note that you do not need the limit of the $f_n$ to be a continuous function!

You should try to construct one such sequence using elementary functions. Try to write down a few examples with some free parameters, and see if you can cook up such a sequence!

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  • $\begingroup$ I feel sure that $\|Tf\|< \|f\| $ when $f\ne 0,$ which is not unusual in infinite-dimensional spaces, but I haven't verified it $\endgroup$ – DanielWainfleet Nov 14 '18 at 13:00
  • $\begingroup$ Yes, but that does not mean that $||\mathcal{T}f||$ could not be arbitrarily close to $||f||$ (which would imply $||\mathcal{T}||_{op}=1$). $\endgroup$ – Angelo Lucia Nov 14 '18 at 18:19
  • $\begingroup$ Seems like your functions aren’t continuous on [0,1]$ but only on $(0,1]$. $\endgroup$ – Shalop Nov 15 '18 at 1:06
  • $\begingroup$ You are right, my mistake. I have removed my example. $\endgroup$ – Angelo Lucia Nov 15 '18 at 3:44
  • $\begingroup$ In reply to your reply to my comment: Yes. And the norm of $T$ is $1$ $\endgroup$ – DanielWainfleet Nov 15 '18 at 6:04

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