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Let $V: C([0,1]) \rightarrow C([0,1])$ be defined by $$ V f(x) = \int\limits_0^x f(t) dt.$$ Show that V has dense range and find the transpose of V.

V has dense range: Since the polynomials are dense in $C([0,1])$, V is dense. The transpose $V^*: C^*([0,1]) \rightarrow C^*([0,1])$ i guess. How does a map between those spaces look like and how do I find it?

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This is not right since all $g$ in the range of $V$ satisfy $g(0) = 0$. Hence the closure of the range of $V$ is not all $C([0,1])$. –  Hans Engler Nov 28 '12 at 18:11

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Operator $V$ doesn't have dense range. Indeed, take $g\in \mathrm{im}(V)$, then for some $f\in C([0,1])^*$ we have $$ g(x)=\int\limits_{0}^{x} f(t)d\lambda(t) $$ where $\lambda$ is a Lebesgue measure on $[0,1]$. Hence $g(0)=0$, and $\Vert 1_{[0,1]}-g\Vert_\infty\geq |1_{[0,1]}(0)-g(0)|=1$. Since $g\in\mathrm{Im}(V)$ is arbitrary we have $1_{[0,1]}\notin\mathrm{cl}(\mathrm{im}(V))$, so image of $V$ is not dense.

By Riesz representation theorem there is isometric isomorphism between $C([0,1])^*$ and $M([0,1])$ the space of signed $\sigma$-additive bounded regular Borel measures. This isomorphism is the following $$ I:M([0,1])\to C([0,1])^*:\mu\mapsto\left(f\mapsto \int\limits_{0}^{1}f(t)d\mu(t)\right) $$ Moreover if for $\varphi\in C([0,1])^*$ we denote $\mu=I^{-1}(\varphi)$ then $$ \varphi(f)=\int\limits_{0}^{1}f(t)d\mu(t) $$ In this case adjoint operator $V^*$ is given by equalities $$ V^*(\varphi)(f)= \varphi(V(f))= \int\limits_{0}^{1}V(f)(x)d\mu(x)= \int\limits_{0}^{1}\int\limits_{0}^{x} f(t)d\lambda(t)d\mu(x)= \int\limits_{0}^{1}\int\limits_{t}^{1} f(t)d\mu(x)d\lambda(t)= $$ $$ \int\limits_{0}^{1}f(t)\int\limits_{t}^{1} d\mu(x)d\lambda(t)= \int\limits_{0}^{1}f(t)\mu([t,1])d\lambda(t)\tag{1} $$ Consider measure $\nu\in M([0,1])$ given by $$ \nu(A)=\int\limits_{A}\mu([t,1])d\lambda(t)\tag{2} $$ where $A$ some Borel set on $[0,1]$. Then $(1)$ can be written as $$ V^*(\varphi)(f)=\int\limits_{0}^{1}f(t)\mu([t,1])d\lambda(t)=I(\nu)(f) $$ Hence $V^*$ maps functional $\varphi\in C^*([0,1])$ to the functional $I(\nu)$.

But if you identify $C([0,1])^*$ with $M([0,1])$ as usually people do the result can be formulated much easier: $V^*$ maps measure $\mu$ to measure $\nu$ given by $(2)$.

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