Why $L^2$ convergence, with Riemann Integral, does not imply pointwise convergence at any point?

Let $$L^2[0,1]$$ be the space of continuous square integrable functions, where we use the Riemann Integral, no Lebesgue allowed. Let $$(f_n)_n$$ be a sequence of continuous functions on $$[0,1]$$ and $$f$$ a continuous functions on $$[0,1]$$. Assume that $$||f_n-f||_2 \rightarrow 0$$. Does it follow that $$f_n(x) \rightarrow f(x)$$ for some $$x \in [0,1]$$? Give a proof or counterexample.

So far, I think that it does follow that $$f_n(x) \rightarrow f(x)$$ for some $$x \in [0,1]$$. Here is why. Suppose for a contradiction that no such point $$x$$ exists. Then we have that $$|f_n(x)-f(x)|> \epsilon$$ for all $$x \in [0,1]$$ and all $$n \in \mathbb{N}$$. Then $$g_n(x)^2=|f_n(x)-f(x)|^2 \geq \epsilon^2$$. It follows that $$\int^{1}_{0}g_n(x)^2dx \geq \epsilon^2$$, so $$\sqrt{\int^{1}_{0}g_n(x)^2dx} \geq \epsilon$$. And we have a contradiction.

Is this reasoning correct?

• No, you may think of the "sliding bump" function as a counterexample. – GNUSupporter 8964民主女神 地下教會 Sep 29 '18 at 22:06
• Could you please point out where my reasoning goes wrong? – G the Stackman Sep 29 '18 at 22:07
• The $\epsilon$ depends on $x$, that's where your reasoning is wrong. – Giuseppe Negro Oct 1 '18 at 9:01

Quick negative response

Sequence of step functions with constant height and contracting width

Take a countable family of sliding bumps $$\{f_{n,k} := \chi_{\Large[(k-1)/2^n,k/2^n]} \mid n \in \Bbb{N}, k\in\{1,\dots,2^n\}\}$$ as a counterexample. $$\int_0^1 f_{n,k}(x)dx = \frac{1}{2^n} \xrightarrow[n\to\infty]{} 0 \implies f_{n,k} \xrightarrow[n\to\infty]{L^2} 0$$ However, $$f_{n,k}(x) \not\to 0$$ everywhere since for each bump of width $$2^{-n}$$, $$k \in \{1,\dots,2^n\}$$ iterates through the whole universe $$[0,1]$$. As a result, $$f_{n,k}(x)$$ "oscillates" between $$0$$ and $$1$$ almost everywhere.

Make our bumps continuous (edited)

As OP points out, the above family of functions are discontinuous, but the essential idea is already there: find a sequence of geometric figures with converging "areas". In the case, a simplest shape that makes $$\{f_{n,k}\}_{n,k}$$ continuous would be $$\triangle$$. $$\begin{array}{rcl} & \Huge\triangle & W(x) = (1-|x|)^+ \\ -1 & 0 & 1 \end{array}$$ Shrink $$W(x)$$ down horizontally by a factor of $$2^{-(n+1)}$$, then translate the peak by $$(2k-1)/2^{n+1}$$ units to the right. $$f_{n,k}(x) = W\left(2^{n+1}\left(x - \frac{2k-1}{2^{n+1}}\right)\right) = W(2^{n+1}x - (2k-1))$$ The integral $$\int_0^1 f_{n,k}(x) dx$$ is just the area of the triangle. Therefore, $$\int_0^1 f_{n,k}(x) dx = \frac12^{n+1}\int_{-1}^1W(x)dx = \frac12^{n+1} \xrightarrow[n\to\infty]{} 0 \implies f_{n,k} \xrightarrow[n\to\infty]{L^2} 0,$$ but $$\{f_{n,k}\}_{n,k}$$ oscillates almost everywhere (on each non-trough points $$[0,1]\setminus\{k/2^n \mid k \in \Bbb{Z}, n \in \Bbb{N}\}$$).

Response to question in the comments

The goal "$$\exists x \in [0,1]: f_n(x) \to f(x)$$" in the attempted proof by contradiction is correctly set up, but the hypothesis "$$\lnot(\exists x \in [0,1]: f_n(x) \to f(x))$$" is incorrectly stated. Things will be clear if we expand everything in $$\forall,\exists$$ quantifiers.

\begin{align} \exists x \in [0,1]: f_n(x) \to f(x) \tag{Goal} \\ \exists x \in [0,1], \forall \epsilon > 0, \exists N \in \Bbb{N}: \forall n \ge N, |f_n(x) - f(x)| < \epsilon \tag{Clearer goal} \\ \forall x \in [0,1], \exists \epsilon > 0: \forall N \in \Bbb{N}, \exists n \ge N: |f_n(x) - f(x)| \ge \epsilon \tag{Negated goal} \\ \exists \epsilon > 0: \forall x \in [0,1], \forall n \in \Bbb{N}, |f_n(x) - f(x)| \ge \epsilon \tag{OP's hypothesis} \end{align}

There're two mistakes in OP's goal:

1. The order of $$\epsilon$$ is moved to the left of $$x$$.
2. $$\forall N \in \Bbb{N}, \exists n \ge N$$ is incorrectly "condensed" into $$\forall n \in \Bbb{N}$$.
This twist in quantifiers only justifies the choice of a subsequence $$(f_{n_k})_{k \in \Bbb{N}}$$ from $$(f_n)_{n \in \Bbb{N}}$$ so that $$\forall k \in \Bbb{N}, |f_{n_k}(x)-f(x)| \ge \epsilon$$.

To avoid such mistakes

1. think of condensations in natural languages for quantifier twist like $$\begin{array}{c|c} \textbf{quantifier} & \textbf{verbal expression} \\\hline \forall N \in \Bbb{N}, \exists n \ge N & \text{infinitely often} \\\hline \exists N \in \Bbb{N}, \forall n \ge N & \text{eventually} \end{array}$$
2. Think about the order of variables in such statements. Never change the order during a negation. If you really need to do so, always appeal to the mathematical definitions/proved results.
3. Write quantifiers for variables before the statements involving those variables. Even though this is often against the order in natural language, you'll see the convenience of doing so when you try to translate them into symbols.
• Thank for the clarification about the use of quantifiers. I believe the function provided as counterexample is not continuous, specifically at the endpoints of the particular intervals. My apologies if I am mistaken. – G the Stackman Oct 1 '18 at 7:21
• Nonetheless, is there a way to turn the function you provided into a continuous function, by connecting the endpoints to value 0 through the use of a suitably chosen linear function, which may depend on n and k? Any further comment is appreciated. – G the Stackman Oct 1 '18 at 8:05
• @GabyBoyAnalysis I've just written down the formal representations for a sequence of triangular bumps so as to the conditions of the question. I'm editing my answer to fix this. Actually, since $\mathcal{C}([0,1],\Bbb{R})$ is complete and separable, we should be able to approximate each real-valued function defined on $[0,1]$ by a sequence of continuous function. The classic diagonal extraction argument will give you a desired sequence across that sequence of constructed sequences. Nonetheless, to avoid overloading readers will results in real analysis, I'll cut my bumps into halves. – GNUSupporter 8964民主女神 地下教會 Oct 1 '18 at 8:15