Attempt:
Sub
the integral becomes :
Another simplier method that i've just thought of :
Suppose the limit exists ,
Let
we have
Perhaps I was wrong , the limit sign is outside the integral
so we cannot put
I am here try to make some corrections ,
Put
By MVT, let
we have
Therefore ,
In fact ,we don't need to put
However . after the corrections , I seem to have chosen a terrible detour . Why I thought in this way ...
BTW , just before the bed , I suddenly think of another method , it is also a simple method :
Integration by parts ,
we just need to prove that the remaining integral tends to zero .
Thanks a lot simplependulum and Krizalid for your help!
Also, I realised that the problem can be generalized.
Notice that
Also, we observe that \forall polynomial P(x):
Now, let f be continuous on [0,1]. Since, the conditions for Weierstrauss Approximation theorem
are satisfied, we can now establish that
Proof :
Given and for we have some polynomial such that
So,
Taking limits we see
That is is near and consequently near .
Since was arbitrary, we are done.