Define $\|p\| := \sup_{[0,1]} |p|$ and consider $T : \mathcal{P} → \mathcal{P}$ as $$T(a_0 + a_1t + · · · + a_kt^k) := a_0 + a_1t +\frac{a_2t^2}{2} + · · · +\frac{a_k}{k}t^k.$$ Show $T$ is a continuous.

I just know $\mathcal{P}$ isn't complete with any norm, so I cannot apply the closed graph theorem and showing that $T$ is bounded seems impossible to me. I appreciate nay help for this problem.


4 Answers 4


$T$ is not continuous. For this it suffices to consider $p_n(t)=1-(2t-1)^{2n}$. Then $\|p_n\|=1$, but $\|Tp_n\|= Tp_n(1)=2\sum_{j=1}^{n}\frac{1}{2j-1}$. Since
$$ \|Tp_n\|\ge S_n:=\sum_{j=1}^{n}\frac 1j \quad\text{and}\quad\lim_{n\to\infty}S_n=+\infty $$ we can take $q_n=p_n/S_n$ and then $\displaystyle\lim_{n\to\infty}\|q_n\|=0$ but $\|Tq_n\|\ge 1$ for all $n$.

  • $\begingroup$ How did you get that expression for $Tp_n(1)$? $\endgroup$ Commented Jun 13, 2017 at 22:20
  • $\begingroup$ I can derive your formula for $Tp_n(1)$ using induction on $n$ and my integral formula for $T$, but I'm curious if there's a more direct way... $\endgroup$ Commented Jun 13, 2017 at 23:02
  • $\begingroup$ Use the equality $\sum_{j=1}^{2n}\binom{2n}{j}\frac{(-2)^j}{j}=\sum_{j=1}^{n}\frac{2}{2j-1}$. $\endgroup$
    – san
    Commented Jun 13, 2017 at 23:04
  • $\begingroup$ The equality can be proved by induction, but probably your method (induction and integral formula) is easier. $\endgroup$
    – san
    Commented Jun 14, 2017 at 12:11

The key observation to understand what's going on here conceptually is that the operator $T$ is given by $$Tp(x)=p(0)+\int_{0}^x\frac{p(t)-p(0)}{t}\, dt.$$ This makes it easy to see that in fact $T$ is unbounded, essentially since the integral $\int_0^1\frac{1}{t}\, dt$ diverges. For instance, for any $\epsilon>0$, we can find (by the Weierstrass approximation theorem) a polynomial $p$ such that $\|p\|\leq 1$, $p(0)=0$, and $p(t)\geq 1-\epsilon$ for all $t\in [\epsilon, 1]$. We then have $$Tp(1)-Tp(\epsilon)=\int_{\epsilon}^1\frac{p(t)}{t}\,dt\geq\int_{\epsilon}^1\frac{1-\epsilon}{t}\,dt=(1-\epsilon)\log(1/\epsilon).$$ This goes to infinity as $\epsilon$ goes to $0$, and so the norms of the $Tp$ also go to infinity. So there is no bound on $\|Tp\|$ for polynomials $p$ such that $\|p\|=1$, and so $T$ is unbounded.


$\newcommand{\norm}[1]{\lVert#1\rVert}$Note: This is not a complete answer but might be a way to approach the answer.

By Stone-Weierstrass, $\mathcal P$'s uniform closure is $\mathrm C[0,1]$. (The space of continuous functions on $[0,1]$)

Define, for each $n \in \mathbb{W}$, $$ T_n\left( \sum_{j=0}^m a_j x^j \right) = a_0 +\sum_{j=1}^n \frac{a_j}{j} x^j $$ where trailing $a_j$'s (Those with indices greater than $m$ but less or equal to $n$) are set to $0$. Then each $T_n$ is bounded via triangle inequality.

If you can somehow show that $T_n$ is a Cauchy sequence in $\operatorname{CHom}[\mathcal P \to \mathrm C[0,1]]$ (the space of continuous linear maps), a theorem in functional analysis would imply that the pointwise limit of $\langle T_n \rangle$ is $T$ and $T$ is bounded in $\norm\cdot$.

  • $\begingroup$ whatever you said it makes sense to me and I agree, but I don't know how to show $\sup_{\|p_m(t)\|\leq 1}\sup_{t\in[0,1]}| \sum_m^n \frac{a_j}{j}x^j|$ is bounded. $\endgroup$
    – Parisina
    Commented Jun 11, 2017 at 1:32
  • $\begingroup$ @Parisina AFAIK it is not. $\endgroup$ Commented Jun 13, 2017 at 2:23

Probably I'm missing something or can it be this simple?

We need to show that $\|Tp\| \to 0$ whenever $\|p\| \to 0$.

Now $Tp$ is continuous and differentiable so it attends $\sup_{t \in [0,1]} |Tp(t)|$ in either $t=0$, $t=1$ or in $t_0\in(0,1)$ such that $(Tp)'(t_0)=0$.

However, $$(Tp)'(t) = a_1 + a_2 t + \cdots + a_k t^{k-1} = \frac{p(t) - a_0}{t}$$ so $(Tp)'(t_0)=0$ implies $p(t_0) = a_0 = p(0)$.

Thus, $$\|Tp\| = \sup_{t \in [0,1]} |Tp(t)| = \max(|p(0)|, |p(1)|) \leq \|p\|$$

When $\|p\| \to 0$ we therefore have $\|Tp\| \to 0$.

  • 2
    $\begingroup$ The inequality $\|Tp\|\le\|p\|$ is not true. Take $p(t)=1-(2t-1)^2=4t-4t^2$, then $\|p\|=1$ but $Tp(t)=4t-2t^2$ and $\|Tp\|=Tp(1)=2>\|p\|$. $\endgroup$
    – san
    Commented Jun 13, 2017 at 21:41
  • $\begingroup$ Or take $p(t)=1-(2t-1)^4=4t(2-6t+8t^2-4t^3)$, then $\|p\|=1$, but $Tp(t)=4t(2-3t+8/3t^2-t^3)$ and $\|Tp\|=Tp(1)=2+2/3$. $\endgroup$
    – san
    Commented Jun 13, 2017 at 21:47
  • $\begingroup$ You need to evaluate $Tp(t)$ at $t=0$, $t=1$, and at the points where the derivative is $0$, but you have instead just evaluated $p(t)$. $\endgroup$ Commented Jun 13, 2017 at 22:55
  • $\begingroup$ What a blunder! Just because $(Tp)'(x) = 0$ where $p(t) = p(0)$ doesn't mean that $Tp$ takes the value $p(0)$ there. $\endgroup$
    – md2perpe
    Commented Jun 14, 2017 at 6:31

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