2楼: Originally posted by xylslyx at 2016-11-16 20:23:21
1)\left \| T \right \|=\left \| t^{2} \right \|_{L_{2}}=\int _{\left }\left |t^{2} \right |^{2}dt=\frac{1}{\sqrt{5}}
2)\left \| T \right \|=\left \| t^{2} \right \|_{L_{\infty }}=1...

3楼: Originally posted by Sunshinelmj at 2016-11-17 09:22:03
取x(t)等于常值函数1，并不一定能保证这是x的范数等于1令Tx取到的最大的吧？不过有启发，我再证下不等式另外半边就行了，谢谢了
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2楼: Originally posted by xylslyx at 2016-11-16 20:23:21
1)\left \| T \right \|=\left \| t^{2} \right \|_{L_{2}}=\int _{\left }\left |t^{2} \right |^{2}dt=\frac{1}{\sqrt{5}}
2)\left \| T \right \|=\left \| t^{2} \right \|_{L_{\infty }}=1...

5楼: Originally posted by xylslyx at 2016-11-17 09:40:08
修正下1）
\left \| T \right \|=\left \| t^{2} \right \|_{L_{2}}=
(\int _{\left }\left | t^{2} \right |^{2}dt)^{\frac{1}{2}}...

6楼: Originally posted by xylslyx at 2016-11-17 09:41:28
修正下1）
\left \| T \right \|=\left \| t^{2} \right \|_{L_{2}}= (\int _{\left }\left | t^{2} \right |^{2}dt)^{\frac{1}{2}}...