TITLE: Showing $T$ is continuous.
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.