Question Number 124415 by snipers237 last updated on 03/Dec/20
![Let a>0, A={f∈C^2 ([0,a],R) , f(0)=f′(0)=0} N_1 (f)= sup{∣f(x)∣+∣f′′(y)∣ ,x,y∈[0,a]} N_2 (f)=sup{∣f(x)+f′′(x)∣ ,x∈[0,a]} Prove that N_1 and N_2 are equivalents norms](Q124415.png)
$${Let}\:{a}>\mathrm{0},\:\:\:{A}=\left\{{f}\in{C}^{\mathrm{2}} \left(\left[\mathrm{0},{a}\right],\mathbb{R}\right)\:,\:{f}\left(\mathrm{0}\right)={f}'\left(\mathrm{0}\right)=\mathrm{0}\right\} \\ $$ $${N}_{\mathrm{1}} \left({f}\right)=\:{sup}\left\{\mid{f}\left({x}\right)\mid+\mid{f}''\left({y}\right)\mid\:\:\:,{x},{y}\in\left[\mathrm{0},{a}\right]\right\} \\ $$ $${N}_{\mathrm{2}} \left({f}\right)={sup}\left\{\mid{f}\left({x}\right)+{f}''\left({x}\right)\mid\:\:\:,{x}\in\left[\mathrm{0},{a}\right]\right\} \\ $$ $${Prove}\:{that}\:{N}_{\mathrm{1}} {and}\:{N}_{\mathrm{2}} \:{are}\:{equivalents}\:{norms} \\ $$
Commented bymnjuly1970 last updated on 03/Dec/20
$$\:\:{Real}\:\:{analysis}\:... \\ $$