On C[0,1] with supremum norm define (Tf)(x)=∫₀ˣx f(t)dt. Let R(T) be its range. P:T is bounded linear. Q:T⁻¹:R(T)→C[0,1] exists and is bounded. Choose truth values.
Linearity is immediate and |Tf(x)|≤x²‖f‖∞≤‖f‖∞. If Tf=0, then ∫₀ˣf=0 for x>0, and differentiation gives f=0, so the inverse on its range exists. But f_n(x)=sin(nx) has norm 1 for large n, whereas Tf_n=x(1−cos nx)/n has norm at most 2/n→0. Thus the inverse is unbounded. **B**.
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