# A bounded linear operator on the space of real polynomials on $[0,1]$ with unbounded inverse operator

I'm trying to do Question 6(ii) on this pdf

Let $$X$$ be the space of real polynomials on $$[0,1]$$ regarded as a subspace of the Banach space $$C[0,1]$$ of continuous functions equipped with the sup norm. For $$k=0,1,2, \ldots$$, let $$m_k$$ be the $$k$$-th monomial, i.e. $$m_k(t)=t^k \quad(t \in[0,1]) .$$ Define $$T: X \rightarrow X$$ by letting $$T m_k=\frac{1}{k+1} m_k$$ and extend $$T$$ to $$X$$ by linearity.
(a) Give an integral expression for $$T x$$ for a general $$x \in X$$ and use this expression to prove that $$T$$ can be extended to a bounded linear operator $$\tilde{T} \in L(C[0,1])$$ with $$\|\tilde{T}\|=1$$.
(b) Prove that $$T: X \rightarrow X$$ is a bijection but that its inverse is not bounded.
(c) Is $$\tilde{T}: C[0,1] \rightarrow C[0,1]$$ still injective? Is it still surjective?

My idea:

(a) $$T:X\to X$$ can be expressed as $$(T f)(0)=f(0),\\ (T f)(t)=\frac{1}{t}\int_0^tf(s)\ ds,\forall t>0$$

$${|t^{-1}\int_0^tf(s)\ ds|}≤{‖f‖}_\sup⇒{‖T‖}≤1$$, also $$T(m_0)=m_0$$, so $${‖T‖}=1$$.

$$X$$ is dense in $$C[0,1]$$, so $$T$$ extends to $$\tilde T:C[0,1]\to C[0,1]$$ with $$‖\tilde T‖=‖T‖=1$$.

(b) Define $$T^{-1}$$ by $$T^{-1}(m_k)=(k+1)m_k$$ and extend it to $$X$$ by linearity.

so $$T$$ has an inverse, so $$T$$ is a bijection.

$$\|T^{-1}(m_k)\|/\|m_k\|=k+1\to\infty$$, so $$T^{-1}$$ is not bounded.

(c) $$\tilde T$$ is injective but not surjective.

$$\tilde T$$ is injective, because we can recover $$f$$ from $$\tilde Tf$$:$$\frac{d}{dt}\bigl(t\cdot\tilde Tf(t)\bigr)=f(t)$$

$$\tilde T$$ is not surjective, because for any $$f\in C[0,1]$$,$$t\cdot\tilde Tf(t)=\int_0^tf(s)\ ds\in C^1[0,1]$$and $$C^1[0,1]$$ is a proper subspace of $$C[0,1]$$.

For example, $$\frac{d}{dt}\left(t|t-\frac12|\right)=(2t-\frac12) \operatorname{sgn}(t-\frac12)$$, so $$t|t-\frac12|\notin C^1[0,1]$$, so $$|t-\frac12|$$ is not in the range of $$\tilde T$$.

• @AnneBauval Ok. Commented Jun 9 at 10:21

• Your initial post was fine (except the lack of any question about your solution). Your recent edit (following Theo's advice, but now deleted) "For every continuous function $$f$$ we have $$\lim_{t\to0}t^{-1}\int_0^tf(s)\mathrm ds=f(0)$$ by mean value theorem which shows $$\tilde T$$ [you meant $$\tilde Tf$$] is continuous at $$0$$ from the right" was useless, if not misleading. The fact that $$\tilde Tf$$ is continuous at $$0$$ is guaranteed by the fundamental theorem of calculus: $$F(t):=\int_0^tf(s)\mathrm ds$$ is differentiable (on the right) at $$0$$ with $$F'(0)=f(0)$$, i.e. $$\lim_{t\to0^+}\tilde Tf(t)=\tilde Tf(0)$$. Moreover, though often invoked when proving the FTC, the MVT or the extreme value theorem are unnecessary. The FTC can be proved directly and elementarily from the definition of continuity.
• Your proof that $$\tilde T$$ is not surjective does not need an explicit counterexample. You could simply notice that (by the FTC again)$$\operatorname{Range}(\tilde T)\subset\{g\in C[0,1]\mid g\text{ is differentiable on }(0,1]\}\subsetneq C[0,1].$$
In part (a), the integral formula is good: we can define: $$(\tilde{T}f)(t) = \begin{cases} \displaystyle{\frac{1}{t}\int_0^t f(s) \, \mathrm{d}t}& \text{if } 0 < x \le 1 \\ f(0) & \text{if } x = 0. \end{cases}$$ This does leave an open question: is $$\tilde{T}f$$ in $$C[0, 1]$$? The fundamental theorem of calculus says $$\tilde{T}f$$ is continuous on $$(0, 1]$$, but a separate argument (possibly involving the mean value theorem) is necessary to show that $$\tilde{T} f$$ is continuous at $$0$$ from the right.
• Typos: $x$ should be $t$ (twice), and $(\tilde Tf)(0)=f(0)$, not $0$ . More importantly, no "separate argument (possibly involving the mean value theorem) is necessary to show that $\tilde Tf$ is continuous at $0$ from the right". This (as well as the fact that $\tilde Tf$ is not only continuous but $C^1$ on $(0,1]$) is guaranteed by the FTC, and the latter can be proved directly from the definition (with no use of the MVT). Commented Jun 9 at 8:30
• @AnneBauval Sure, I agree, that's one way to prove continuity. I interpreted the OP's argument as multiplying the continuous function $\int_0^t f(s) \, \mathrm{d}s$ by the continuous function $\frac{1}{t}$, neglecting the $0$. I would still say that an extra argument, such as the one you articulated in your answer, is still necessary, even if it is straightforward (at least, at this level). Commented Jun 9 at 9:22