# Answer to Question #16022 in Functional Analysis for Ciaran Duffy

Question #16022

Suppose f is a function such that

f(x) = x^2 x element Q,

0 x element R − Q.

Is f continuous at c = 0? Is f differentiable at c = 0?

f(x) = x^2 x element Q,

0 x element R − Q.

Is f continuous at c = 0? Is f differentiable at c = 0?

Expert's answer

1) The function is continuous at c=0.

Indeed, fix any

epsilon>0

and put

delta = epsilon.

We can assume that

epsilon<1.

Then for any x with

|x|<delta

we have that

|f(x)| < epsilon.

Indeed, if x is an element of Q, then

|f(x)| =

|x^2| < |x| < epsilon, since |x|<1.

if x is an element of R-Q,

then

|f(x)| = 0 < epsilon.

Thus f is continuous at

c=0.

2) f is also differentiable at c=0, in the sense that

lim_{x->0} ( f(x)-f(0) ) / x = 0.

Indeed, fix

epsilon>0

and put

delta = epsilon.

Then for any x with

|x|<delta

we have that

| (f(x)-f(0)) / x | <

epsilon.

Indeed, if x belongs to Q-0, then

| [ f(x)-f(0) ] / x

| = | (x^2 - 0) /x | = |x| < epsilon

and if x belongs to R-Q, then

f(x)=0, and

| (f(x)-f(0)) / x | = | (0 - 0)/x | = 0 <

epsilon

Thus

lim_{x->0} [ f(x)-f(0) ] / x = 0.

3)

On the other hand f is not continuous and not differentiable at any other point

c<>0.

Indeed, if c<>0, then take two sequences

(a_i) and

(b_i) converging to c such that each

a_i is raitonal, i.e. belogns to

Q

and each

b_i is irrational, i.e. belogns to R-Q.

Then

lim_{i -> infinity} f(a_i) = lim_{i -> infinity} a_i^2 =

c^2,

while

lim_{i -> infinity} f(b_i) = lim_{i -> infinity} 0 = 0

<> c^2.

So f is not continuous at c, and therefore it is not

differentaible at c as well.

Indeed, fix any

epsilon>0

and put

delta = epsilon.

We can assume that

epsilon<1.

Then for any x with

|x|<delta

we have that

|f(x)| < epsilon.

Indeed, if x is an element of Q, then

|f(x)| =

|x^2| < |x| < epsilon, since |x|<1.

if x is an element of R-Q,

then

|f(x)| = 0 < epsilon.

Thus f is continuous at

c=0.

2) f is also differentiable at c=0, in the sense that

lim_{x->0} ( f(x)-f(0) ) / x = 0.

Indeed, fix

epsilon>0

and put

delta = epsilon.

Then for any x with

|x|<delta

we have that

| (f(x)-f(0)) / x | <

epsilon.

Indeed, if x belongs to Q-0, then

| [ f(x)-f(0) ] / x

| = | (x^2 - 0) /x | = |x| < epsilon

and if x belongs to R-Q, then

f(x)=0, and

| (f(x)-f(0)) / x | = | (0 - 0)/x | = 0 <

epsilon

Thus

lim_{x->0} [ f(x)-f(0) ] / x = 0.

3)

On the other hand f is not continuous and not differentiable at any other point

c<>0.

Indeed, if c<>0, then take two sequences

(a_i) and

(b_i) converging to c such that each

a_i is raitonal, i.e. belogns to

Q

and each

b_i is irrational, i.e. belogns to R-Q.

Then

lim_{i -> infinity} f(a_i) = lim_{i -> infinity} a_i^2 =

c^2,

while

lim_{i -> infinity} f(b_i) = lim_{i -> infinity} 0 = 0

<> c^2.

So f is not continuous at c, and therefore it is not

differentaible at c as well.

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