Bounded operator but not Compact

compact-operatorsfunctional-analysisoperator-theoryunbounded-operators

Let $T : (C([-1,1]),||.||_{\infty}) \rightarrow (C([-1,1]),||.||_{\infty}) $

Such as : $(Tf)(x)=\frac{1}{2}(f(x)+f(-x))$ . For all $f\in C([-1,1])$

Why $T$ isn't Compact ?

I tried to use the sequence $f_n(x)=x^n$. For $x\in [-1,1]$.

But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.

Best Answer

Hint: Instead, try $$ f_n(x) = x^{2n}, $$ and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.

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