Is limit function of continuous even functions always continuous?

Question

Limit function of Uniform Continuous functions (pointwise converge) is always continuous.

It is the fact.

Limit function of continuous functions is not always continuous (am I right?)

However, for the case that limit function of continuous even functions, what can do we conclude it?

In other words, limit function of continuous even functions is always continuous even not in the case of uniform convergence?

I am pondering this subject for long time, but it is very confusing me.

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Actually, I have this problem.

I am having hard time to solve this problems. $C$ denotes set of functions, $f:[-1,1] \to \mathbb{R}$ and $f$ is continuous. please help me with this.

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Let $C_{e}([-1, 1],\mathbb{R})$ denote the set of even function in $C([-1,1],\mathbb{R})$.

[a] Show that $C_e$ is closed and not dense in $C$.

[b] Show that the even polynomials are dense in $C_e$, but not in $C$.

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The part of solution is,

@qyong,

Since it's been a while, I'm assuming you're still stuck on the problem. I am assuming you have endowed $C([-1,1])$ with the uniform continuity metric for ease. Let's start with a lemma, from which, as @Ian suggested, the solutions should follow nicely:

Lemma: Let $f_i \in C([-a,a])$. Given a sequence of even, real-valued, continuous functions $f_i : [-a,a] \to \mathbb{R}$ that converge to $f$, $f$ must also be even.

Proof: Assume the hypothesis, that we have a sequence $f_{i}(x)$ of continuous, even, real-valued functions that converge uniformly to some $f(x)$ for $x \in [-a,a]$. Assume $f(x)$ is not even. Then we know that for sufficiently large $i \in \mathbb{N}^+$, $|f_{i}(x) \to f(x)| < \epsilon$ for some $\epsilon > 0$. From which, we also have $|f_{i}(-x) \to f(-x)| < \epsilon$ as well. But evenness of the $f_{i}$ also tells us that $f_{i}(-x) = f_{i}(x)$, so that $|f_{i}(x) \to f(-x)| < \epsilon$. But since $f$ is not even, $f(x) \neq f(-x)$, which contradicts convergence of the $f_i$. QED

From this, we have:

[a]. $C_{e}$ contains all of it's limit points, since any $f_{i} \in C_{e}([-1,1], \mathbb{R})$ converges to an even $f$, which is also in $C_{e}([-1,1], \mathbb{R})$. Thus $C_e$ is closed.

[b]. Let $f \in C([-1,1], \mathbb{R})$. We can apply the Weierstrass Approximation Theorem since $[-1,1]$ is closed and bounded, i.e. for some even (again, by hypothesis) polynomial $P(x)$, we have $|P(x) - f(x)| < \epsilon$. So by our lemma above, $f$ must be even, i.e. $f \in C_{e}([-1,1], \mathbb{R})$. Thus if we let $P_{e}$ be the set of all even polynomials, by the above argument in conjunction with Weierstrass's Theorem, we have that $\overline{P_e} = C_e$, so it is dense in $C_e$, but $\overline{C_e} \neq C$ so the set of even polynomials cannot be dense in $C$.

By @Alex

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However, how about, not in the uniform continuous?????????

I think this solution is not perfect........

Answer 1

The answer is no, and the standard counterexample almost works.

The standard counterexample is $f_n(x) = x^n$ on $[0,1]$. Simply extending this counterexample, and keeping only the even functions, gives you the functions $$f_n:[-1, 1]\to \mathbb R\\ f_n(x) = x^{2n}$$

Which converges to a non-continuous function.

Answer 2

The pointwise limit of a sequence of continuous functions $(f_n)$ may not be continuous you're right. Take the example of the sequence of functions defined on $[0,1]$ by $f_n(x)=x^n$. The sequence converges pointwise to the function $f$ equal to $0$ on $[0,1)$ and to $1$ at $x=1$. Obviously $f$ is discontinuous at $1$.

$f_n(x) = \vert x \vert^n$ is even and continuous and the limit is not continuous.

To be even has little thing to do with continuity. Imagine that a sequence of continuous functions $(f_n)$ is defined on the interval $[0,a]$ with $a > 0$. If the functions converges pointwise to a discontinuous function, the sequence $(\overline{f}_n)$ defined on $[-a,a]$ by $\overline{f}_n(x)=f_n(x)$ for $x \ge 0$ and $\overline{f}_n(x)=f_n(-x)$ for $x \le 0$ is even and is also discontinuous.