Loomis and Sternberg Problem 1.58

Null Space of $D$:

If $f \in W$ s.t. $f(x)=c \in \mathbb{R}$, then $Df \in V$ is such that $Df(x)=0$ for any $x$ which is the additive identity of $V$ so $f \in N(D)$.

If $Df=f_0^V$, the additive identity of $V$, then $Df=Dg$ where $g(x)=\int_a^x f_0^V(t) \ dt =\int_a^x 0 \ dt = c_1 \in \mathbb{R}$. Since $Df=Dg$, $f=g+h$ where $h$ is some constant function so $f$ is also some constant function.

Thus $N(D)=\{f:f(x) = c\} \subset W$. $\square$

$D$ is surjective:

Take any $f \in V$. Let $F \in W$ be $F:x \mapsto \int_a^x f(t) \ dt$. Then $DF=f$ so $D$ is surjective. $\square$

$T$ is injective:

$T(f_0^V) = F$ where $F:x \mapsto \int_a^x f_0^V \ dt = 0$ which is the additive identity of $\ W$.

Assume $T(g)=F$ is the additive identity of $W$. Then $F(x)=\int_a^x g(t) \ dt = 0$ then $DF(x)=0=g(x)$ which is the additive identity of $V$.

Therefore $N(T)=\{f_0^V\}$ and $T$ is injective. $\square$

Not sure how to find the range of $\ T$. Would really appreciate some hints. Thanks!

Solution 1:

Actually, $T$ is not surjective. Its range is the strict subspace $$ \mathcal{C}^1_{a \mapsto 0} [a, b] := \big\{F \in \mathcal{C}^1 [a, b] \ :\ F(a) = 0\big\} $$ of $\mathcal{C}^1 [a, b]$. That is, its range only comprises of continuously differentiable functions whose values are $0$ at $a$.

• Indeed, if $F \in T\big(\mathcal{C}^0 [a, b]\big)$, then there is $f \in \mathcal{C}^0 [a, b]$ such that $F(x) = T(f)(x) = \int_a^x f(t)\ dt$; but then $F(a) = \int_a^a f(t)\ dt = 0$ showing $F \in \mathcal{C}^1_{a \mapsto 0} [a, b]$.

• And if $F \in \mathcal{C}^1_{a \mapsto 0} [a, b]$, then $F \in \mathcal{C}^1 [a, b]$ and $F(a) = 0$. Thus its derivative $f := F' \in \mathcal{C}^0 [a, b]$ and so by FTC $$ T(f)(x) = \int_a^x f(t)\ dt = \int_a^x F'(t)\ dt = F(x) - F(a) = F(x) - 0 = F(x) $$ i.e. $T(f) = F$. Hence $F \in T\big(\mathcal{C}^0 [a, b]\big)$.

You can also see the non-surjectivity of $T$ in the fact that $T$ is not the left inverse of $D$ despite $D$ being the left inverse of $T$. Indeed, consider the function $I_{42}(x) = x + 42$ in $\mathcal{C}^1 [0, 1]$. $D(I_{42})(x) = 1$ in $\mathcal{C}^0 [0, 1]$ and so $(T \circ D)(I_{42})(x) = \int_0^x 1 \ dt = x$ in $\mathcal{C}^1 [0, 1]$, which is clearly not the function $I_{42}$ we started with.