Continuity of a Max Function on [0,1] X [0,1]

Let f(x,y) be a real valued continuous function defined on the unit square [0,1] X [0,1]. Prove g(x)=max{f(x,y) : y in [0,1]} is continuous. --- Can we treat g(x) as a composite function that maps R^2 --> R ?

Real Analysis : Limits Proof

Prove sqrt(x+1) - sqrt(x) goes to 0 as x --> infinity Please see the attached file for the fully formatted problems.

Real Analysis : Lipschitz Condition

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Real Analysis : Lipschitz Condition

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Real Functions of Real Variable, Continuity

Let f:R->R satisfy |f(t)-f(x)|<=|t-x|^2 for any t,x. Prove that (f) is constant.

Connected and Continuous

There are a couple of concepts I need clarification for: 1) If a set has no interior points, then is it necessarily closed? Isn't the empty set considered open? 2) If the graph of f: R -> R is connected, does it have to be continuous? In just Real -> Real aren't these definitions equivalent? Thanks.

Bijection is Homeomorphism

Let f : M -> N be a continuous bijection. M is compact. Show that f is a homeomorphism. Isn't a homeomorphism by definition a bijection? And since M is compact, will it not be true that N will be compact too?

Differentiable Function

Assume f : R -> R is differentiable and there exists an L < 1 such that for each x in R, f'(x) < L. Prove that there exists a unique z in R such that f (z) = z.

Difference Quotient : Limits and Differentiable Functions

Assume f:(-1,1) --> R and f'(0) exists. If a_n , b_n -> 0 as n->infinity, define the difference quotient: D_n = ( f(b_n) - f(a_n) ) / ( b_n - a_n). a) Prove lim [n -> infinity] D_n = f'(0) under each condition below: (i) a_n < 0 < b_n . (ii) 0 < a_n < b_n and (b_n) / (b_n - a_n) <= M (iii) f'(x) exists and is contin ...continues

Proofs : Riemann Integrable Functions

Let RI be the set of functions that are Riemann Integrable. Disprove with a counterexample or prove the following true. (a) f in RI implies |f| in RI (b) |f| in RI implies f in RI (c) f in RI and 0 < c <= |f(x)| forall x implies 1/f in RI (d) f in RI implies f^2 in RI (e) f^2 in RI implies f in RI (f) f^ ...continues