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 $\tilde T$.

  • $\begingroup$ @AnneBauval Ok. $\endgroup$
    – hbghlyj
    Commented Jun 9 at 10:21

2 Answers 2


Though you removed your temporary "Is my solution to part (c) correct?", which was even not precise enough for a solution verification post, let me comment your solution.

  • 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)$$\operatorname{Range}(\tilde T)\subset\{g\in C[0,1]\mid g\text{ is differentiable on }(0,1]\}\subsetneq C[0,1].$$
  • A more educated (though admittedly extremely overkill) argument for this non-surjectivity was the open mapping theorem. I guess that the ultimate goal of the whole exercise was to illustrate it.

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.

Parts (b) and (c) are fine, despite any doubts.

  • $\begingroup$ 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). $\endgroup$ Commented Jun 9 at 8:30
  • 1
    $\begingroup$ @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). $\endgroup$ Commented Jun 9 at 9:22

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