Suppose f g D rightarrow R are both continuous at a in D Sho

Suppose f, g: D rightarrow R are both continuous at a in D. Show from the definitions that h(x) = max(f(x), g(x)) is continuous at a. You will need the fact that for any e > 0, and any L, M, max(L - e, M - e) - max(L, M) - e, max(L + e, M + e) = max(L, M) + e.

Solution

Let > 0 be given and aD.

We shall show that h is continuous at a.

Since, f and g are both continuous at aD, so there exist ₁>0 and ₂ >0 such that for any xD

|x-a|<₁ implies |f(x)-f(a)|< and |x-a|<₂ implies |g(x)-g(a)|<

Choose =minimum{₁,₂}, then

for any xD, |x-a|< implies |f(x)-f(a)|<  and  |x-a|< implies |g(x)-g(a)|<    ..(i)

Now, without any loss of generality let us assume that f(a)g(a), then h(a)=f(a).

Consider, for xD and |x-a|<,  |h(x) - h(a)|= |max(f(x),g(x)) - max(f(a),g(a))| = |max(f(x),g(x)) - f(a)| ..(ii)

Now, h(x)=f(x) or g(x)

if h(x)=f(x) then |h(x) - h(a)|= |f(x) - f(a)|< [from (i) & (ii)]

therefore, h is continuous at a.

if h(x)=g(x) then |h(x) - h(a)|= |g(x) - f(a)| [from (ii)]

< f(x) f(a) g(x) f(a) g(x) g(a) < [since f(a)g(a)]

which implies, |g(x) - f(a)|<, therefore |h(x) - h(a)|= |g(x) - f(a)|<

Hence we conclude that h is continuous at a.