are continuous). Then we also know that the above function g is continuously differentiable on [a, b]. For that purpose, we will first show that

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1 Derivatives and Integrals Now suppose f : [a, b] [c, d] IR is continuous. It is easy to see that for each fixed x [a, b] the function f x (y) = f(x, y) is continuous on [c, d] (note that f x is sometimes used to denote the partial derivative of f w.r.t. x). In particular f x R[c, d] and hence the function g(x) = d c f(x, y)dy is well defined. Using uniform convergence of f(x n, y) to f(x 0, y) on [c, d] (or using the definition of limit), we see that d lim n) = n c f(x 0, y)dy for any sequence (x n ) in [a, b] such that lim n x n = x 0. It then follows that g is a continuous function on [a, b]. Next suppose f C 1 ([a, b] [c, d]). That is, f and all its first directional derivatives are continuous (it is equivalent as requiring both x and y are continuous). Then we also know that the above function g is continuously differentiable on [a, b]. For that purpose, we will first show that g(x n ) g(x 0 ) d lim = n x n x 0 c x (x 0, y)dy for any sequence (x n ) [a, b], x n x 0 and lim n x n = x 0. Again, the above can be shown by showing that f(xn,y) f(x 0,y) x n x 0 converges uniformly to x (x 0, y) on [c, d]. To do this, use Mean value theorem and uniform convergence (or using the definition of limit). 75

2 S, 2016 What about improper integrals? We will just demonstrate it for the case [0, ). Suppose f : [a, b] [0, ) IR is continuous such that the improper integral 0 f(x, y)dy converges uniformly on [a, b]. Then again the function g(x) = 0 f(x, y)dy is a continuous function on [a, b]. Finally, we have the following theorem for derivative and improper integrals. simplicity, we only discuss only one type of improper integral here. Theorem Suppose f C 1 ([a, b] [c, )), and both the improper integrals c f(x, y)dy and c on [a, b] and (x, y)dy converge uniformly on [a, b], then g is continuously differentiable x g (x) = c (x, y)dy. x For What about the derivative of the function x c f(x, y)dy?

3 3110S, Implicit function theorem and inverse function theorem Let f : IR IR be such that f (c) 0. If f is continuous on a neighborhood of c, then f will be of constant sign on a (possibly smaller) neighborhood of c. For simplicity, we may assume that this neighborhood is an open interval I. It then follows from the mean value theorem that f(x) f(y) for all x y, x, y I. Hence f is 1-1 on that interval. Moreover, the range of the interval must be also an open interval J (why?). Hence, we can define an inverse function for f on J. This is called the inverse function theorem. Indeed, the above is true for a continuously differentiable function f : D IR n IR n. If the linear transform (f)(c) (c D, D is open) is invertible, then there exist an open set U (in D) containing c and an open set V containing f(c) such that f is 1-1 from U onto V. It is then possible to define an inverse function f 1 : V U. With the help of chain rule, we see that f 1 is also differentiable and Id n n = f f 1 (x) (f 1 ) x.

4 S, 2016 Closely related to inverse function theorem is the implicit function theorem. Let f : IR 2 IR be continuously differentiable such that x 2 (c) 0 c = (c 1, c 2 ). Then there exists an open interval I containing c 1 and a differentiable function g : I IR such that g(c 1 ) = c 2 and f(x, g(x)) = f(c) for all x I. Moreover, g (x) = x 1 / x 2 on I. Proof. Indeed, if f : IR n+1 IR is continuously differentiable such that x n+1 (c) 0 c = (c 1,, c n+1 ), then there exist an open set U IR n containing (c 1, c 2,, c n ) and a differentiable function g : U IR such that g(c 1,, c n ) = c n+1 and f(x, g(x)) = f(c) for all x U. Moreover, xi g(x) = g x i = x i / x n+1 on U. Proof: Consider the function f(x) = (x, x,, f(x)). Then f : IR n+1 IR n+1 and f c is invertible. Remark Of course, there is also an implicit function theorem for f : IR n+m IR m being continuously differentiable such that the derivative of h = (f 1, f 2,, f m ) at c is invertible. Note that conversely, implicit function theorem implies inverse function theorem.

5 3110S, Taylor s formula By letting g(t) = f(c+t(z c)) we are able to produce a Taylor s formula using partial derivatives. Extremum Similar to one variable functions, if a function has a local extremum at a point c (of an open set), then f(c) must be the zero vector (if it exists). Moreover, Theorem Let f : IR 2 IR be twice continuously differentiable on an open set U and c U. Suppose f(c) = 0. Denote A = x1 x 1 f(c), B = x1 x 2 f(c) and C = x2 x 2 f(c) and 2 f(c) be the two by two matrix ( xi x j f(c)) i,j=1,2. Then (i) If det ( 2 f(c)) > 0 and A > 0 (and hence C > 0), then f has a local minimum at c. (ii) If det ( 2 f(c)) > 0 and A < 0 (and hence C < 0), then f has a local maximum at c. (iii) If det ( 2 f(c)) < 0, then f has a saddle point at c.

6 S, 2016 Remark (i) On IR n, a similar theorem for above will be if x T ( 2 f(c))x 0 with equality only at x = 0 (usually known as ( 2 f(c)) being positive definite), then c is a local minimum. Similarly, if the quadratic form is negative definite, then c is a local maximum. (ii) If c is a local extremum of f, then Proof. (a) d dt t=0f(c + t v) = 0 if it exists. (b) d2 dt t=0f(c + t v) 0 (if it exists) when c is a local maximum. 2 (c) d2 dt t=0f(c + t v) 0 (if it exists) when c is a local minimum. 2 (d) In particular x 2 i f 0 at local maximum and x 2 i f 0 at local minimum. (e) If d 2 dt 2 t=0f(c + t v) < 0 for all nonnezero v, then c is a local maximum. d 2 (f) If dt t=0f(c + t v) > 0 for all nonnezero v, then c is a local minimum. 2 For (ii), just consider the function g(t) = f(c+t v). For (i), it is a consequence of Taylor s formula.

7 3110S, Lagrange multiplier It is usually necessary to find extremum of functions under constraint. For example, let f, g : D IR n R be (at least) continuous. We want to find the extremum of f on {x D : g(x) = 0}. First, we will find all possible points of local extremums. To do that, we have a useful method: Lagrange multiplier. The method of Lagrange multiplier is based on the following fact: If f has a local extremum at c subject to the constraint g(x) = 0, then f c is a scalar multiple of g c if g c 0. In general, under the constraints, g i (x) = 0 for i = 1,, m we can conclude that f c = m i=1 λ i g i c for some λ 1,, λ m IR. Proof. To see that, first note that if γ : ( ε, ε) D is differentiable such that γ(0) = c and g i (γ(t)) = 0 for all i, then g i c γ (0) = 0. Since c is a local extremum, we have 0 = d dt t=0f(γ(t)) = f c γ (0). Next, we will use the implicit function theorem to show that if g i c v = 0, for all i and v 0, then there exists differentiable γ : ( ε, ε) D such that γ(0) = c, γ (0) = v and g i (γ(t)) = 0 for all i. It then follows that f c v = 0. It is now clear that f c is in the linear span of { g i c : i = 1,, m}.

8 S, 2016 Riemann Integrals on higher dimension Let R be a parallelepiped with edges parallel to axes, that is R = [a 1, b 1 ] [a n, b n ]. For simplicity, we will say R is a closed interval in IR n. P is said to be a partition of R if P = {R j : j = 1,, N}, N N, R j s are closed interval in IR n such that R = N j=1r j and they intercept each other only at the boundary and not the interior. We could then define upper and lower sum U(P, f), L(P, f) just as before and so are upper integral and lower integral of f on R. Again, we say a function f is Riemann integral on R if it is bounded and its upper integral coincide with its lower integral. It is then easy to see that continuous functions are Riemann integrable. Moreover, one can again show that (but its proof will be more complicated): R fdx = lim R(P, f, ξ) = lim f(ξj )vol(r j ). P 0 P 0 We will take P be the maximum of lengths of closed intervals R j in the partition. It is then clear that sum and product of integrable functions are integrable and Riemann integral is linear. An important class of Riemann integrable functions are Characteristic functions of domains D with D having measure zero. It is integrable as it is continuous on both D and D c. Fubini s theorem It is clear that if f is continuous, then b d a c f(x, y)dydx = d b c a f(x, y)dxdy. The theorem follows from the fact that linear combinations of products of single variable functions are (uniformly) dense in C(R) (the set of continuous functions on R = [a, b] [c, d]). That is given any ε > 0, there exist continuous functions functions h i C[a, b], g i C[c, d], i = 1,, N, N N such that f(x, y) N i=1 h i (x)g i (y) R < ε.

9 3110S, Indeed, Fubini s theorem holds for any Riemann integrable function f such that f x (y) = f(x, y) is integrable on [c, d] for each x [a, b] and f y (x) = f(x, y) is Riemann integrable for each y [c, d]. Fubini s theorem for improper integral We will just discuss a typical example [0, ) [0, ) here. Theorem Let f R([0, N] [0, N]) for all N and for all ε > 0, there exists N > 0 such that R V S U f(x, y)dxdy < ε for all R > S > N, V > U > N. Suppose also that f x (y) and f y (x) are both integrable on all finite intervals. Then 0 0 f(x, y)dxdy = 0 0 f(x, y)dydx.