find adjoint operator of an operator A

Question

How to find adjoint operator of an operator A $$A \in B(C^1[0,1], C[0,1])$$ $$ (Ax)(t) = x'(t)?$$ In answer : for any functional $f_y$ originated by function $y \in BV_0[0,1]:A(f'_y) = g_z$, where functional $g_z$ originated by couple of function $z(t) = y(t)$ and number zero.

Have no idea how to find. Can you help me with this?

@SmileyCraft for C[0,1] :$\parallel x \parallel = max | x(t)|$, where $t \in [0,1]$ — Gera Slanova, Jan 08 '19 at 21:50
@SmileyCraft $\parallel x\parallel =max|x'(t)|$, where $t\in[0,1]$ — Gera Slanova, Jan 08 '19 at 21:56
Then $A$ is not bounded. Consider $f_n(t)=t^n$. Then $|f_n|=1$, but $|Af_n|=n$, since $f_n'(t)=nt^{n-1}$. — SmileyCraft, Jan 08 '19 at 21:58
@SmileyCraft sorry, I had made a mistake. for x $\in C^1[0,1]$ $\parallel x \parallel = max \parallel x ^{(k)}\parallel_{C[0,1]}$ where $0\leqslant k \leqslant 1$ — Gera Slanova, Jan 08 '19 at 22:09

score 3 · Answer 1 · answered Jan 08 '19 at 21:32

3

Take $\mu\in C[0,1]^*$. By definition, $A^*\mu\in C^1[0,1]^*$ is given by $$ (A^*\mu)(f)=\mu(Af)=\mu(f')=\int_{[0,1]}\,f'\,d\mu. $$

answered Jan 08 '19 at 21:32

Martin Argerami

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find adjoint operator of an operator A

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