# Self-adjoint operator and inner product

I am wondering whether there is a way to make sense of self adjointness of an operator on $C[0,1]$ without resorting to the inner product of $L^2[0,1]$.

I am not referring to concrete alternative ways to check, but rather to whether it is even sensible to ask whether an operator is self-adjoint if one does not have an inner product.

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The adjoint of an operator $T: X \to X$ is $T^*: X^* \to X^*$ given by $(T^* f)(x) = f(T x)$. Now in the case of $X = C[0,1]$ you have an embedding $J$ of $X$ in $X^*$ given by $(Jx)(y) = \int_0^1 x(t) y(t)\ dt$, so you can ask for $T^* \circ J = J \circ T$. Of course you could say this is introducing an inner product. Moreover, there are many other possible embeddings $J$, each of which would give a different notion of "self-adjointness".

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