To make our life easier we are going to pick Ts = 1, which means that, my god, n is equal to pi, so general derivation is given into bulk. We make a couple of claims now. The space of pi-bandlimited functions is a Hilbert space. The set of functions 5n of t, which are simply succinct functions shifted to location n, form an optimal basis for that Hilbert space. And if x of t is pi-bandlimited, then the sequence of samples xn is a sufficient representation. In other words, if we have the samples xn, we can reconstruct xt perfectly and that's the essence of the sampling theorem. So the space of pi bandlimited functions is a vector space. Actually it's a subspace of L2 of R. This we shall not prove, but it is intuitive because if you add two bandlimited function, the result will be a bandlimited function. So that's the closure of the vector space. The inner product is a standard inner product in L2 of R and the completeness issue, which is of course, a trickier one is more delicate. And it we'll not address it here. Just remember the definition of the inner product, and the fact that the convolution can be written as an inner product. Just taking care that we do the change of variable t minus tau, in one of the two elements of the convolution. Let us prove that the sync function and its shifts by integer is an autonormal basis for the pie bandlimited space. For this we need to show that the inner product between phi n and phi m is going to be equal to 1 when n is equal to m and 0 otherwise. So let's write this inner product between phi n and phi m, we can see that's the same as between phi 0 and shapes of t- n and t- m. Then we use the fact that sinc function is symmetric in this argument. So, (t- n) can be changed into (m- t). Therefore, we have the integral of sinc(t- n) sinc (m -t) dt. This we can factor as the convolution of two sinc functions shifted by (m- n) Now we can use a convolution theorem knowing that a fourier transform of a sinc function is a rect function. So sinc convulse with sinc at the location, m- n is 1 over 2 pi's integral of rect square. And then e to the j omega m- n d omega. This is simply the integral over minus pi to pi of this function that we have seen before. And this is equal to 1, when m is equal to n, 0 otherwise. With this we have shown that sinc functions are to each other when shifted by Integers and the sinc function itself is off no more. So we have an optimal set of vectors in bl of pi, this set of optimal vectors can be shown to be an optimal basis for bandlimited functions.