# Value of operator norm when $\mathcal{T}f(x)=\int^{x}_{0} f(t)dt$

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.

• Sorry, had to make another edit due to a typo... – Joe Man Analysis Nov 14 '18 at 2:21
• see volterra integral operator – qbert Nov 14 '18 at 3:08
• Maybe, take $f_n=1_{[0,1/n]}$. Compute the norm of $f_n$ and of $Tf_n$. – Shalop Nov 14 '18 at 3:17
• @Shalop. That's not a sequence of continuous functions. – md2perpe Nov 14 '18 at 9:32
• @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. – Shalop Nov 14 '18 at 12:43

## 2 Answers

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 $$(Tf_n)(x)\geq (Tf_n)(\frac {1}{n+1})=1.$$ $$\text {So }\quad \|Tf_n\|\geq \int_{1/(n+1)}^1 (Tf_n)(x)dx\geq \int_{1/(n+1)}^1 1\cdot dx=$$ $$=1-\frac {1}{n+1}.$$

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

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!

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