Lecture 45

Welcome back we are discussing about the Fourier transform on R and we got inversion formula as well as Plancherel theorem and then how we related the Fourier transform via Poisson formula to a periodic function Fourier series and so now in next few lectures we will be seeing some application of this Fourier transform on R. So one of the very important result in real analysis is the Weierstrass approximation theorem that says that if f is a continuous function on a closed bounded interval a b then there exist a sequence of polynomial which converges to f uniformly. So today we are going to see an alternate proof which is pretty much close to the classical proof of Weierstrass himself by using by exploiting the properties of good kernels we had developed during our journey in Fourier transform on R. So what is Weierstrass theorem? Let be a continuous function on closed bounded interval a b and then for every epsilon positive there exist a polynomial p such that supremum of x varies over a b mod of p x minus of f of x this is less than epsilon. Okay, so now coming to the proof we have real line here it is a and b so f is compactly supported on a b so I can get a m and minus m such that a b is contained in minus m to m so let a b is contained in minus m to m. Okay, so define g of x this is equal to 0 if x is outside m minus m to m and let us see that I have a continuous function on a b it looks like this is our f now and g is equal to f of x when x belongs to a b and continuous otherwise that means what one can talk of we have many choices of g, g is going to be 0 here and on here straight away I can just make this line a b and this joining the line so this is what and inside a b g is going to agree with f. function on R, g is continuous on R with compact support. Okay, now take K delta be a good kernel K delta in precisely we can take to be 1 by square root of delta e to the power minus pi x square by delta. Now if I take the Gaussian and then dilate it we have used it quite often in our previous lectures then K lambda K delta is a good kernel this family is a good kernel for delta greater than 0. So that means g is a continuous function and what we have seen is that we have seen earlier that g convolution with K delta this converges to g uniformly. Okay, so what does this mean so for epsilon positive there exist a delta not greater than 0 such that supremum with g x minus g convolution of K delta of x this is less than epsilon this is less than K delta not is less than epsilon that is what the meaning of converging uniformly. Now look at for each delta what does this K delta will look like this is equal to 1 by square root of delta not and e to the power minus pi x square by delta which we can write we know the definition which is equal to minus pi x square by delta not whole to the power n by n factorial this is the series representation of the Gaussian. Now we know that this series converges therefore the tail of this series is going to be small so which will say that for there will exist a n sub not such that K delta minus of r x this is less than epsilon by some constant 4 m we have taken where b is the supremum of mod g where what is r? r is the first n sub 0 term so this is equal to 1 by square root of delta not summation over n equal to 0 to n sub not minus pi x square by delta not to the power n by n factorial that means this is what if I am subtracting this r x then what I am having basically the tail of this series given by this because this series is convergent therefore I can say that tail is small. Now what we can see is that because we are chasing for a candidate which is going to be a polynomial so mod of g convolution of the K delta not of x minus g convolution of r x this if I write down the definition of that so and push the mod inside this would be lesser equal to minus m to m because g is supported on minus m to m g of y and then mod of K delta not of x minus of y minus r of x minus of y dy just writing the definition of the convolution. So now if I pull out the sup norm of g over here and I know that this is less than epsilon by 4 mb so this is going to be b times epsilon by 4 mb into 2 m so this is small therefore now g of x minus g convolution of r of x this is equal to g of x minus g convolution of K delta not of x plus g convolution of K delta not of x minus of r x which is lesser equal to mod of g of x minus g convolution of K delta not of x plus mod of g convolution of K delta not of x minus g convolution of r of x this is g convolution of r of x. Now this already we have seen that this is less than epsilon and this is already less than epsilon I can make this the initial state this is less than epsilon by 2 so in this case this is epsilon by 2 epsilon by 2 this is equal to epsilon. Now if we can if we can real proof that g convolution of r is a polynomial in x then we will be done no we have started with f so now if we are putting the x in a b then this is f of x minus g convolution of r of x is less than epsilon. So what we want to show is that if we can show that this is a polynomial then we are done so what is g convolution of r of x let us write it down explicitly this is minus m to m g of y r of x minus of y dy. Now as you can see that by definition of r this is equal to minus m to m g of y then 1 by square root of delta not summation over n equal to 0 to some finite number n sub naught minus 1 to the power n pi to the power n x to the power 2 x minus of y to the power x square is 2 n and then divided by n factorial and then there is a factor of delta to the power n this is what we are going to get. So now as this all these are finite sum so what one can also this is independent I pull this minus 1 to the power n pi to the power n by n factorial and then integral minus m to m g of y then x minus of y 2 n then this is dy now which is equal to 1 by square root of delta not minus m to m g of y. Now I we can use the binomial theorem binomial expansion so k is equal to 0 to 2 n this is 2 n choose k and x to the power k and minus of y to the power 2 n minus of k dy. So you take that sum out so this is square root of delta not n equal to 0 to n naught minus 1 to the power n and then k from 0 to 2 n 2 n choose k then x to the power k and then minus m to m g of y and minus y to the power 2 n minus of k dy and this is a constant depending on n and k. So therefore what we got is that this is k is equal from 0 to 2 n and then 1 over square root of delta not summation over n from 0 to n sub naught minus 1 to the power n by delta not to the power n n factorial then this is 2 n choose k for each k and then this C n k and then x x to the power k. So this is clearly a polynomial in x so r is g convolution of r at x is a polynomial in x and this is what we wanted because what we have shown earlier that f of x for x belongs to f of x minus g convolution of r of x is less than epsilon and this is a polynomial this is the required polynomial which does the job for us if f is a continuous function on a closed and bounded interval a b. Okay so now as the next application we will be using we will be kind of seeing the mathematical version of Heisenberg uncertainty principle which has one of the main feature in the quantum mechanics however we will not be discussing any physics over here. So what it says is that the you know in quantum mechanics this is if I am looking for a position and momentum then the position is not talked about exactly where this is placed it is in a probabilistic sense it is given by the integral and everything is given by a state function function which is which in our case we will take the state function to be in the Schwarz space with that integral mod of psi x square dx is equal to 1. So then converting the mathematical version of the uncertainty principle which says that you cannot determine the exact position of a particle and the momentum. So what it turns out to be a mathematical inequality in terms of the Fourier transform suppose psi belongs to the Schwarz space with integral minus infinity to infinity mod of dx is equal then mod of psi x square this is basically given by the position psi hat of xi square d xi which is greater or equal to 1 by 16 pi square. So what it says is that both psi and psi hat this integral cannot be very small because it is always has a the product of them always has a lower bound which means if you are taking this integral to be very small then in order to satisfy this inequality this has to be large and vice versa and equality holds if and only if psi of x is nothing but some scalar multiple of the dilation of the Gaussian where b is greater than 0 and mod a square this is equal to square root of 2 b by as a matter of fact in fact we can have we have if you take the variance with minus infinity to infinity deviation x naught square psi x square dx into square d xi this is greater or equal to 1 by 16 pi square okay. So let us prove this so what we know we have been given it that 1 is equal to minus of infinity to infinity mod of psi x square dx and psi is a Schwarz class function so it is differentiable and has nice decay so this we can write it as by integration by parts x into mod psi x square and then this is minus infinity to infinity then minus of x into d by dx of psi x square is psi x into psi x bar dx this is integration by parts so first one because psi in the Schwarz space it kills all the polynomial so at infinity when we are going because of the rapid decay for psi this term is going to be 0 and so we are left with minus infinity to infinity only with this quantity and this is x then this is psi prime of x psi bar of x plus psi of x and then psi prime of x bar dx. So now this will imply one this is a positive number so therefore if I push the mod this is going to be minus infinity to infinity mod of x and then this is mod of psi x psi prime of x psi bar x and this is also modulus is going to be same so we can get 2 times of this of dx that is what easily we could get. So now we are in a position to apply the Cauchy-Schwarz inequality so this is less or equal to 2 times minus infinity to infinity mod of x square mod of psi x square dx whole to the power 1 over 2 into minus infinity to infinity mod of psi prime of x square dx whole to the power 1 over 2. So now this is equal to 2 times minus infinity to infinity mod x square mod psi x square as it is and then what we know is by Plancherel theorem this integral this is going to be minus of infinity to infinity mod of psi prime hat at xi square d xi. Now this psi prime hat we know that this is 2 pi i xi mod square and then this is psi hat of xi mod square d xi this is what so this is 4 pi square is and with a factor of half if I am taking it into 2 pi and then this is minus infinity to infinity mod xi square mod of psi hat of xi square d xi whole to the power 1 half okay. Now so that is what what we have got this implies that minus infinity to infinity mod of x square psi x square dx multiplied by minus infinity to infinity xi square mod of psi hat of xi square d xi this is square 1 by 4 pi square this is greater or equal to 1 by 16 pi square. Now the question what we would like to address is that when this inequality becomes equality. So the inequality is going to become the equality if and only if the Cauchy Schwarz inequality is going to become the equality because everything in between steps they are going to be equality. So now the Cauchy Schwarz is equality occurs if and only if so there are two function what we are taking psi prime of x which is equal to some beta times x psi x in the Cauchy Schwarz inequality. Now if we solve this this is the ordinary differential equation. So now what we can get is that psi of x is equal to a e to the power some beta x square by 2 that is what is going to be and because this psi has to be in Schwarz space therefore beta has to be negative that is what so now psi of x is equal to a e to the power minus b x square that is what is the required one the equality is going to occur. And if you take this then obviously you can calculate because we know how precisely how to calculate the Fourier transform of this Gaussian and that is what is going to give us the equality. Thank you.