 Welcome to lecture 34 on measure and integration. In this lecture, we look at some special spaces which are constructed on measure spaces. These spaces play important roles in topics like functional analysis, harmonic analysis and so on. So, we will be studying today's space is called L p spaces. So, we will fix a real number between 0 and infinity p. So, p is a real number between 0 and infinity and we look at the space called L p mu which is also written as L p of x s mu depending on whether we want to emphasize the underlying measure space or not. If it is clear from the context, what is the underlying set s x and the sigma algebra s, we will just write this space as L p mu. So, this is the space of all complex valued s measurable functions on the space x such that integral of the absolute value of the function f raise to power p d mu is finite. We recall in the previous lectures, we had defined the notion of function which is complex valued on a set x and which is s measurable and we also defined the notion of its integral. So, if we take a function f which is complex valued such that the absolute value of this function raise to power p which is a non-negative measurable function. If that is integrable of mod f to the power d mu is finite then we say the function f is p th power integrable and the collection of all p th power integrable functions on the measure space x s mu is denoted by either L lower p x s mu or just L p mu. So, L p mu is the space of all p th power integrable functions on x s mu and today we are going to study properties of this set L p x s mu. So, the first observation we want to make is that the space L p x s mu can be treated as a vector space over the complex numbers under the addition and scalar multiplication of functions. So, let us observe how that is done. So, we have got L p of mu. So, that is the space of all p th power integrable functions. So, we want to show that if we define f plus g x to be f x plus g x for every x and alpha f x to be equal to alpha times f of x. Then under this operation of addition and scalar multiplication L p mu is a vector space. So, for that we will have to show that alpha times f is a function in L p of mu whenever f is a function in L p of mu. So, let us check that. So, alpha belongs to c and f belongs to L p of mu. Let us look at we want to check alpha times f is in L p or not. So, we have to look at the absolute value of alpha f raised to power p and we have to show this is a integrable function and its integral is finite. But it is obvious this is equal to mod alpha to the power p and mod f to the power p by the property of the absolute value. So, thus implies that integral of mod alpha f to the power p d mu is equal to integral of mod alpha to the power p into the product mod f to the power p. But integral of a scalar times a function is nothing but the scalar times the integral of the function. So, by that property this is and because f belongs to L p of mu. So, this is finite. So, if implies that alpha f belongs to L p of mu. So, scalar multiple of functions in L p are again functions in L p. Let us look at the second property namely the addition. So, let us take two functions f and g belonging to L p of mu. We want to show that mod of f plus g raised to power p is integral is finite. But let us observe. So, we want to show that this d mu is finite. So, let f and g belong to L p of mu to show we want to show that f and g belong to L p of mu. That means integral of mod this to the power p d mu is finite. So, this is what we have to show. So, let us look at the function mod of f plus g. We know that this is less than or equal to mod f plus mod g by the absolute value of by the absolute triangle inequality of the absolute value. So, this to the power p is less than or equal to this is the power p. And now let us observe that the right hand side mod f plus mod g is less than or equal to 2 times the maximum value of mod f and mod g. Because mod f will be less than the maximum of mod f mod g and mod g also is less than maximum of mod f and mod g. So, mod f plus mod g is less than maximum twice the maximum of mod f and mod g. So, this raise to power p. But that is same as we can take this 2 to the power p out. So, this is less than or equal to 2 to the power p and the maximum of 2 numbers is always less than or equal to. So, it is less than or equal to the sum of those. So, let us first observe that this is actually equal to 2 to the power p maximum of mod f to the power p and mod g to the power p. And now this is less than or equal to 2 to the power p mod of f to the power p plus mod of g to the power p. Because maximum of 2 numbers is always less than or equal to the sum of the 2 numbers. So, what we get is that mod f plus f plus g mod to the power p is less than 2 to the power p mod f p plus mod f g. So, that gives us the inequality. So, integrating both sides, we will get that integral of mod f plus g raise to power p will be less than 2 to the power p times integral of mod f to the power p d mu plus integral of mod g to the power p d mu. And both of them being finite. So, this is a finite quantity. So, that implies that whenever f and g belong to l p of mu, then mod of f plus g also or f or f plus g, the function f plus g also belongs to l p of mu. So, that proves that proves the fact that l p is a vector space over the field of complex numbers. Next, let us define for a function f belonging to l p of x mu for l p. What is called the p th norm of the function? Because f belongs to l p, so the integral of mod f to the power d mu is a finite number. It is a finite non-negative number. So, we can take its p th root. So, 1 over p of this number is called the p th norm of the function f. So, norm f p, so the lower index p indicates that we are taking the p th power of the function to integrate and then taking the p th root of the integral. So, this is called the p th norm of f. We want to show that this p th norm as the following properties namely norm of p is always bigger than or equal to 0. That is obvious, because we are integrating a non-negative function. So, integral of mod f to the power p is always non-negative. And if the function is 0 almost everywhere, then of course, the integral is 0. So, the norm is equal to 0. Conversely, if the norm of the function is equal to 0, if the p th norm is equal to 0, that means integral of mod f to the power p is 0. And being a non-negative function that implies f of x must be 0 almost everywhere. So, this property 1 is something similar to what we have done for when p is equal to 1 for the space of integrable functions. So, the p th norm of the function is always bigger than or equal to 0. And it is equal to 0 if and only if f of x is equal to 0. The second property that the norm of the function alpha times f is same as the absolute value of alpha times the norm of f. So, this double bar also indicates the absolute value of the scalar alpha. That is again obvious, because once we take the p th norm of this function. So, let us verify this fact, namely for alpha belonging to C and f belonging to L p, if we look at the norm of alpha times f. So, that is equal to look at the function alpha f, take the power p, integrate out with respect to mu and look at the 1 p th root of that. But that is equal to mod alpha f to the power p is same as mod alpha to the power p integral mod f to the power p d mu raise to power 1 by p. And now when we open it out, so mod alpha to the power p raise to power 1 over p is mod alpha into integral of mod f to the power p d mu raise to power 1 over p, which is nothing but the norm. So, this integral is nothing but the p th norm. So, that proves the property namely that alpha times f p th norm is equal to mod alpha times the p th norm. And the third property we want to prove is that the function f plus g, which we know if f and g belong to L p, then the function f plus g belongs to L p. So, we want to claim that this satisfies the triangle inequality, namely norm of f plus g is less than or equal to norm of f plus norm of g. For p equal to 1, this property was obvious. We had that followed basically, because mod of f plus g is less than or equal to mod f plus mod g. So, integrating both sides, we got integral of mod f plus g is less than or equal to integral of mod f plus integral of mod g. So, that means the norm of f plus g is less than or equal to norm of f plus norm of g. So, for p equal to 1, this is obvious, but for p not equal to 1, we need to do some more calculations to prove this result. So, we need first of all, we will first prove it for the cases when p is strictly bigger than 1. So, we will be looking at the real number p, which is strictly bigger than 1 and of course, less than infinity. So, for such p, we need a lemma, which says that for every non-negative real numbers a and b, if we fix t between 0 and 1, then the following inequality holds namely a raise to power t, b raise to power 1 minus t is less than or equal to t times a plus 1 minus t times b. If you look carefully, for t equal to 1 by 2, this is just saying that the geometric mean is less than or equal to arithmetic mean. So, this is generalization of the standard inequality that the geometric mean is always less than or equal to arithmetic mean. So, what we are saying is for any real number t, positive real number t between 0 and 1, for non-negative real numbers a and b, a raise to power t, b raise to power 1 minus t is less than or equal to t times a plus 1 minus t times b. Of course, if to prove this, let us observe that, if a either a is equal to 0 or b equal to 0, then the left hand side is equal to 0 and the right hand side also is equal to 0. So, in that case it is a equality. So, if either a is 0 or b is 0, both sides are equal to 0 and there is nothing to prove. So, let us assume that both a and b are not equal to 0. So, in that case, let us observe that, proving this inequality that a to the power t, b to the power 1 minus t is less than or equal to a to the power t plus 1 minus t times b is same as, we can rewrite this inequality as this b raise to power 1 minus t is same as b times b divided by b raise to power t. So, that b raise to power t in the denominator we have commended with a raise to power t. So, write this as a by b raise to power t and that b which was to the power 1, we shift it to the other side. So, that goes to a divided by b times t plus 1 minus t times b divided by b which is equal to 1. So, the required inequality is same as proving that a divided by b raise to power t is less than or equal to t times a divided by b plus 1 minus t. Now, let us just rewrite that. So, this is same as saying bring everything on all the terms on one side. So, that is same as saying 1 minus t plus t times a by b minus a by b times t is always bigger than or equal to 0. So, we have to show that this is always bigger than or equal to 0. So, let us put this quantity a by b as x. So, we have to show that for every x bigger than 0, we want to show that 1 minus t plus t x minus x to the power t is always bigger than or equal to 0. Now, realize the left hand side is a function of x and we want to show that function of x is always a non negative function. So, one way of showing that would be that we look at this function f of x and realize that the value of this function at the point x is equal to 1 is equal to 0. So, showing that this inequality holds is showing that f of x is always bigger than or equal to f of 1. So, let us write that. So, let us write the function f of x equal to 1 minus. So, let us write the function f of x is equal to 1 minus t times plus t times x plus x to the minus x to the power t. Then let us calculate f of 1, which is equal to 1 minus t plus t x, t x is 1. So, that is 1 minus x is equal to 1 minus t x to the power t. So, t is x is equal to 1. So, that is 1 to the power t that is equal to. So, that is t times x. So, that is t and this is equal to 1. So, this is equal to 0. So, we want to show that f of x is bigger than or equal to 0, which is f of 1 for every x. So, that sort of indicates that we should try to show that this function f of x has got a minimum value at the point x is equal to 1. So, claim. So, there is required inequality will be through if we can show that f of x has minimum at x is equal to 1. So, let us analyze and that is done by using the tools of calculus. So, let us use tools of calculus to analyze the maximum and minimum of the function f of x, which is equal to 1 minus t times plus t of x minus x to the power t. So, we realize that this function is differentiable everywhere and we calculate the derivative of this function. So, that is equal to 1 minus t is a constant and derivative of t x with respect to x that is equal to t minus t times x to the power t minus 1. So, to calculate the critical points f dash x equal to 0 implies. So, this is t 1 minus x to the power t minus 1 equal to 0 and that implies x is equal to 1. So, the function has a critical point at x is equal to 1 and to analyze whether it is a maximum or a minima. Let us look at apply the second derivative test. So, from here we will have f double dash of x will be equal to t is a constant. So, minus t into t minus 1 into x to the power t minus 2. So, at f dash at 1 that is equal to minus t into t minus 1 and t being a number between 0 and 1 minus t is negative t minus 1 is negative. So, this is bigger than 0. So, second derivative at the critical point 1 is bigger than 0. So, implies that x is equal to 1 is a point of local minima. So, that proves the required property that the function has a local minimum and hence the property that. So, that proves the property that the function f of x has a local minimum at the point x is equal to 1 and hence the required inequality namely f of x is bigger than or equal to 0 is bigger than 1 holds. So, this proves the lemma namely for any two non-negative real numbers a and b and for a real number t fixed between 0 and 1 a raise to power t times b raise to power 1 minus t is less than or equal to t times a plus 1 minus t times b and we will be using this lemma to prove another equality for our spaces L p spaces which is called holders inequality. So, let us state what is called holders inequality. Holders inequality says that for real numbers p and q p bigger than 1 and q bigger than 1 such that 1 over p plus 1 over q is equal to 1 for such numbers p and q if I take a function f which is in L p and look at a function g which is in L q then f times g is a function which is in L 1 and integral of f g the absolute value of f and g product is less than or equal to the integral of mod f to the power p raise to power 1 over p and mod of g raise to power q raise to power 1 over q. So, that essentially says that the function f g is integrable. So, it has the L 1 norm. So, one can state the holders inequality as saying that the L 1 norm of f g is less than or equal to the product of p th norm of f and the k th norm of g. So, that is called holders inequality. So, let us prove this holders inequality. So, we have got two functions f and g. So, f belonging to L p and g belonging to L q where 1 over p plus 1 over q is equal to 1 and we want to show that the norm f g is less than or equal to the p th norm of f and the k th norm of g. So, this is the inequality we want to prove. Let us observe first of all note. Let us call this if norm of f equal to 0 or norm of g is equal to 0. If either of these two quantities are equal to 0, what will that mean? So, norm of f equal to 0 implies integral of mod f to the power p d mu is equal to 0 and that will imply that the function f x equal to 0 almost everywhere and that will imply that the function f g equal to 0 almost everywhere. So, if norm of f is equal to 0, then the function f g is equal to 0 almost everywhere. So, this l 1 integral of this is equal to l 1 norm of the function f into g is also equal to 0. So, both sides will be equal to 0. Similarly, if norm of g is equal to 0, then again both sides of the inequality will be 0 and this will be a equality. So, the required inequality holds as an equality if either of norm f or norm g is equal to 0. So, let us suppose that norm of f p is not equal to 0 and norm of g is also not equal to 0. So, we are now going to apply the lemma. So, let us consider the case, special case when t is equal to 1 over p and the number a is equal to mod f divided by norm f to the power p, norm f the whole thing to the power p and b is absolute value of g divided by norm of g. So, we are going to apply the lemma namely a raised to power t, b raised to power 1 minus t is less than or equal to t times a plus 1 minus t times b with t equal to 1 over p a equal to this number mod f divided by norm of f whole to the power p, which is defined because norm f is not 0 and similarly b equal to norm of absolute value of g divided by norm of g whole thing raised to power q. So, when we do that, so t raised to power 1 over t, so this is t a, so mod f whole to the power p, so that gives you mod f, so that gives us mod f divided by norm of f and b raised to power 1 minus t is equal to 1 over p and note 1 minus t is 1 minus 1 over p, which is equal to 1 over q, so that gives you norm of absolute value of g divided by the norm of g. So, that is the left hand side of the inequality is less than or equal to t, which is 1 over p times a, so a is mod f divided by norm of f p whole raised to power p and similarly 1 minus t, which is 1 over q and b, which is nothing but mod g divided by norm of g to the power 1 by q. So, this is the application of that inequality and let us now simplify this a bit further and now observe that, so mod f mod g divided by norm f norm g is less than or equal to 1 over p times this quantity. So, let us integrate both sides with respect to mu, so that will give you integral of mod f g d mu divided by norm of f mu divided by norm of f mu because these are just constants is less than or equal to 1 over p, which is a scalar and mod f to the power p d mu mod f to the power integrate both sides, so that gives you the norm of f to the power p and so this is also norm f to the power p, so that gives you 1 over p plus 1 over q and that is equal to 1, so that gives you, so integrating both sides we get that norm f g is less than or equal to norm of f p norm of g q, so that is called holders inequality. So, let us go back to revise this again, what is called holders inequality holders inequality says that, if p is bigger than 1 and q is bigger than 1, so that 1 over p plus 1 over q is equal to 1, then for function f belonging to l 1 l p and g belonging to l q the product f into g is in l 1 and its integral is less than or equal to the norm p th norm of f into p th norm of g. So, once again let us go to the proof, so we write a as the norm of f and b as the norm of g, so if either a is 0 or b is 0, then either the function f will be 0 or the function g will be 0 and both sides of the inequality will be equal to 0, so the required claim will hold. So, let us suppose a is not 0 and b is not 0, so then we can divide by a and b, so let us write a to be norm f, a to be absolute value of f divided by capital A, what is capital A? Recall capital A is the norm of f, so whole thing raise to power p and similarly b is g x absolute value divided by the norm of g raise to power q and t equal to 1 over p. So, apply the lemma, so the lemma will give us that a raise to power t is 1 over p, so mod f over a and then into g times into b times b raise to power 1 minus t will give you g divided by b is less than or equal to t times, so t is 1 over p polite times a, so that is a plus 1 over q times b, so that is now integrate both sides with respect to mu, so integral of f g with respect to mu is less than or equal to integral of f p, which is nothing but a to the power p, so that cancels out, so integrate while integrate this cancels out, this cancels out, so this is 1 over p plus 1 over q and this a b you can take it on this side, so that gives you that integral of f x g x is less than or equal to a times b, which is a norm of f times norm of g, so this is called holder's inequality. Using this inequality we will prove another inequality, which is called Minkowski's inequality, which is essentially the triangle inequality for the l p norm, so it says that if f and g are in l p then of course we have already shown that f plus g is in l p and the claim is that the norm of f plus g is less than or equal to norm of f plus norm of g, so this we will prove using holder's inequality, so let us start the proof. So, for p equal to 1 we have already analyzed the proof and seen it is easy, so let us assume p is strictly bigger than 1, so when p is bigger than 1, let us look at the function, we know f plus g belongs to l p, so look at the function f plus g raise to power p minus 1, so note that we have the special relation between p and q, namely 1 over p plus 1 over q is equal to 1 and that is same as saying the number p is also written as p equal to p minus 1 times q. So, that is falling from this is from the inequality, so this is coming from the equality that, so let us just recall that we have seen that 1 over p plus 1 over q is equal to 1, so that is cross multiply, so that is q plus p is equal to p q and that says that p is equal to p q minus q, so that is same as saying from here q is common, so p minus 1 times q, so that is one observation that if p and q have the relation 1 over p plus 1 over q is equal to 1, then p can be written as this, so once that is true let us look at the function which is, so consider the function which is mod f plus g raise to power p minus 1, we want to claim that this belongs to l q, so for that because the reason for that because the reason is mod f plus g raise to power p minus 1, so we want to now raise it to the power q integral d mu, so what is that, so that is equal to integral of f plus g p minus 1 into q that we have already seen p minus 1 into q is p, so that is equal to p d mu and that is finite, so that proves that if f and g are in l p, then mod f plus g raise to power p minus 1 is in l q, so this observation will be used soon, so let us write consider the holders in equality with the functions mod f into f plus g raise to power p minus 1 and mod g into mod of f plus g raise to power p minus 1, note f is in l p, this function is in l p and f plus g raise to power p minus 1 is in l q, so by holders in equality this function is integrable and its integral is less than or equal to norm of f plus norm of this function similarly g is in l p and f plus g raise to power p minus 1 is in l q, so once again this product will be in l 1 and holders in equality will apply, so we start the proof by observing that mod f times mod of f plus g raise to power p minus 1 with this function being in l q, so this is in l p, this is in l q, so the product is l 1, so that will be less than or equal to the p th norm of f that is the p th norm of f plus the k th norm of this function mod f plus g raise to power p minus 1, so what is the k th norm, so it is f plus g the function is to the power p minus 1 for the k th norm raise it to the power q the whole thing raise to power 1 minus q and that is same as this p minus 1, so this is equal to this one we have already seen it is equal to p, so this thing is norm of f plus g raise to power p raise to power 1 over q, so this integral is nothing but norm of f plus g raise to power p by q, so we get the inequality using holders inequality namely the product of the function mod f and mod f plus g raise to power p minus 1 is less than the p th norm of f times the p th norm of f plus g raise to power p by q, a similar application of holders inequality to the second function it will give us that the mod of g times mod of f plus g raise to power p minus 1 is less than the norm of g times the p th norm of the second function which is nothing but mod f plus g raise to power p by q, so now let us use these two to calculate the norm of f plus g, so to calculate the norm of f plus g let us raise it to the power p, so the p th power of the norm of f plus g is nothing but integral of mod f plus g raise to power p and now the trick is that this power p we write it as p into p minus 1, so this is equal to p minus 1. This integral is nothing but integral of f plus g raise to power 1 into the same thing raise to power p minus 1, so this number mod f plus g absolute value is written as mod f plus g and times mod f plus g raise to power p minus 1 and now the first absolute value f plus g we use triangle inequality this is less than or equal to mod f plus mod g, so this is by triangle inequality from here and now this integral can be split into two integrals, so then p th power of the p th norm of f plus g will be less than or equal to integral of mod f times mod f plus g raise to power p minus 1 plus integral of mod g times that and then we will use the earlier obtained bounds, so we have got this is less than or equal to this integral plus this integral and this here we are using that holders equality the bound we have obtained, so this integral mod f times mod f plus g raise to power p minus 1 is less than or equal to the p th norm of f and the k th norm of this function and the k th norm of this function is mod f plus g raise to power norm raise to power p by q and similarly the second term is less than norm of g times the norm of f plus g raise to power p by q, now this norm of f plus g raise to power p by q is common, so let us take it out, so this is less than or equal to norm f plus norm g times norm of f plus g raise to power p minus p by q, so on the left hand side if you recall we had the norm raise to power p and now on the right hand side we have got the norm the one term is norm raise to power p by q, so take it on the other side, so we get norm of f plus g raise to power p minus p by q is less than or equal to norm of f plus norm of g and now using the fact that 1 over p plus 1 over q is equal to 1 realize that this number is nothing but p common, so 1 minus 1 over q that is equal to 1 over p, so that cancels, so this number is equal to 1, so we get what is called holders inequality which is also same as a triangle inequality for the p th norm namely norm of f plus g is less than or equal to norm of f plus norm of g, so that is called Minskowski's inequality, so what we have shown is the following that for the space of p th power integrable functions is it is a vector space 1 and secondly for every function f in this space we can define its l p norm which is the integral of absolute value of the function mod f to the power p the whole thing raise to power 1 over p and we have just now shown it has the three properties namely the norm l p norm of a function is bigger than or equal to 0 and it is equal to 0 if and only if the function is 0 almost everywhere and second property that the l p norm of alpha times f is equal to absolute value of alpha times the norm of p and the third the triangle inequality namely norm of f plus g is less than or equal to norm of f plus norm of g, so as in the case of l 1 let us identify functions which are equal almost everywhere, so for f and g in l p if we identify functions which are equal almost everywhere then we observe that the norms of these two functions are same, so norm of a function f is equal to the norm of a function g if f and g are equal almost everywhere, so if we identify functions which are equal almost everywhere we will get their norms to be same, so we will do that, so in l p we will not distinguish between functions which are equal to 0 almost everywhere, so with that understanding let us define the distance between two functions in l p to be d f g equal to norm of f minus g with respect to p, so once we do that this becomes a metric on l p, so d f g is a metric on l p because d f g equal to 0 if and only if f is equal to g almost everywhere and we have identified functions which are equal almost everywhere, so this becomes a metric and it is called the l p metric on l p metric on the l p space and like l 1 one claims what is called Ries-Fischer theorem, namely the space l p is a complete metric space in the under this metric l which is called l p metric and the proof of this theorem is verbatim same as the proof for saying that l 1 of a b is complete, so we will just sketch the proof and ask you to verify the steps which will also help you to revise the earlier proof and if you still have difficulty you can refer to the textbook, so the steps of the proof are as follows let us take a Cauchy sequence in l p, we want to show that this Cauchy sequence is convergent in l p, so for that the first step is because this sequence is a Cauchy sequence we can select integers positive integers n 1 less than n 2 and less than n k such that the consecutive difference between f n k plus 1 and f n k is less than 2 to the power minus k and because Cauchy says the elements of the sequence are going to come closer and closer as you go to infinity, so using that we can select inductively integers n 1 less than n 2 less than n k and so on such that the difference between f n k plus 1 and f n k is the norm is less than 2 to the power minus k. Once that is done let us define g k to be the function mod f n 1 plus the sums 1 to k of the differences f n j plus 1 minus f n j for every k we define this sequence. Then this g k is a function claim is that if you take the power of this to the power p of this, then this function g k to the power p with l 1 norm is actually we should be defining this modification we should make f n 1 to the power p plus f n 1 f n j plus 1 minus f n mod to the power p. So, this one will require that you define, so there is a sprint here this should be mod f n to the power p plus mod of f n j plus 1 minus f n j absolute value to the power p, so that is g k. So, once that is done if you integrate both sides then we get that this is less than or equal to this power p is here, so integral of the power p, so that is gives the norm of the function f n 1 to the power p and this to the power p that is finite and hence this g k's will belong to g k's will belong to l p space. Now, as a consequence of monotone convergence theorem the series form if we define g x then using series form of the dominated convergence theorem one shows that this function g also is in l p and hence deduce that this integral of mod f n 1, so g k this is finite and hence the function f x which is the partial sum which is the sum of the series f n 1 x minus this is exist almost everywhere and so as a consequence we will get this f n k because the partial sums of just f n k that converges to f of x and now an application of dominated convergence theorem again gives you that f belongs to l p and the integral of mod f n k plus 1 minus f to the power p converges to 0, so norm converges to. So, the proof is essentially the same as that of l 1, so I have just outlined the steps try to prove the steps yourself and convince that it is. So, this is what is called Ries Fischer theorem, so this is we have proved it for p bigger than 1 and less than infinity one can ask the question that what happens for this number p between 0 and 1 for p between 0 and 1 when you can define this space is called l p spaces and show they are metric spaces. However, there is a problem that when you if you try to define the norm as integral of mod f to the power p raise to power 1 over p that does not satisfy the training inequality. So, one can define a metric d f g to be for p between 0 and 1, we can define the metric to be the integral of the absolute value f minus g to the power p, but that does not really help it becomes a metric one can show it is complete also, but the problem comes that f going to norm if you define that norm it is not a norm. So, integral of mod f to the power p is not a norm, so the triangle inequality fails. So, that is why for p between 0 and 1 these spaces are not very interesting for applications point of view. Another observation now one can also define what are called l infinity spaces namely you look at you say a function f defined on x is essentially bounded if there exists a real number m such that the measure of the set where mod f x is bigger than m is equal to 0. So, you collect together functions which are essentially bounded and call them as l infinity. So, one can show that l infinity is a vector space and if one defines for a function f in l infinity what is called the l infinity norm to the infinimum of these constants m such that the measure of the set where f x is bigger than m is equal to 0. Then one can show that this indeed is a norm on l infinity it is called the essential supremum of f and one shows that if you identify functions f and g to be same if they equal almost everywhere then this f going to l infinity is a norm. So, that gives a metric now namely the distance between f and g to be norm of f minus g and this indeed is a metric and one can show that l infinity becomes a complete metric space like l p for p bigger than or equal to 1. So, these are examples of metric spaces which arise out of measure theory. So, let me the importance of this lies in the following fact. So, whenever one is given a vector space and a norm is defined on it that space is called a norm linear space and every norm give rise to a metric. So, the metric being the distance between f and g to be the distance the norm of f minus g. So, that gives a norm and if under that norm under that induced metric induced by the norm if this vector space becomes complete metric space then the space is called a Banach space. So, l p space is p bigger than or equal to 1 including l infinity are examples of Banach spaces. And this p equal to 2 is a very special space then one can even define the notion of angle and relate distance and angle. So, l 2 give examples of a Hilbert space. So, today we have looked at l p spaces which give very important examples of norm linear spaces in fact Banach spaces and also l 2 gives example of a Hilbert space. All these spaces play an important role in the subject of functional analysis and in harmonic analysis. So, if you go for higher studies you will come across these spaces again in your studies. So, let me stop here today. Thank you.