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Fedor Petrov
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Your first condition yields $$|p(c)|=|p(1)|+\int_{1}^c |p'(x)|dx:=\|p(x)\|.$$ All linear functionals on a finite-dimensional space are bounded, so if $\deg p\leqslant n$, we get $|p(0)|\leqslant C_n \|p(x)\|$ for certain $C_n$. Thus, if $b>C_n$, the second condition is not achievable.

Your first condition yields $$|p(c)|=|p(1)|+\int_{1}^c |p'(x)|dx:=\|p(x)\|.$$ All linear functionals on a finite-dimensional space are bounded, so if $\deg p\leqslant n$, we get $|p(0)|\leqslant C_n \|p(x)\|$ for certain $C_n$. Thus if $b>C_n$, the second condition is not achievable.

Your first condition yields $$|p(c)|=|p(1)|+\int_{1}^c |p'(x)|dx:=\|p(x)\|.$$ All linear functionals on a finite-dimensional space are bounded, so if $\deg p\leqslant n$, we get $|p(0)|\leqslant C_n \|p(x)\|$ for certain $C_n$. Thus, if $b>C_n$, the second condition is not achievable.

Source Link
Fedor Petrov
  • 108.8k
  • 9
  • 264
  • 459

Your first condition yields $$|p(c)|=|p(1)|+\int_{1}^c |p'(x)|dx:=\|p(x)\|.$$ All linear functionals on a finite-dimensional space are bounded, so if $\deg p\leqslant n$, we get $|p(0)|\leqslant C_n \|p(x)\|$ for certain $C_n$. Thus if $b>C_n$, the second condition is not achievable.