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Question:

I'm studying a sequence of functions $$f_n(x) = \sqrt{x^2 + \frac{1}{n}}$$ defined on the domain $[-1,1]$ and I'm trying to understand their behavior as $n$ approaches infinity.

Context and Motivation:

This problem came up while I was studying real analysis and the concept of uniform convergence. I understand that for a sequence of functions to converge uniformly to a limiting function, the speed of convergence must be independent of the choice of $x$ in the domain.

I'm interested in seeing how this plays out for the given sequence of functions, especially since they involve a square root, which can often lead to interesting behavior.

My Attempts:

I've tried to find the pointwise limit of the sequence by taking the limit as $n$ goes to infinity, which gives me the function $$f(x) = |x|$$.

However, I'm unsure if the convergence is uniform. I've also noticed that each $f_n$ is differentiable, but I'm not sure if the limiting function $f$ is differentiable on $[-1,1]$ due to the absolute value.

Specific Questions:

  1. Does the sequence $f_n(x)$ converge uniformly to $f(x)$ on $[-1,1]$?
  2. Is the limiting function $f(x)$ differentiable on $[-1,1]$?
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1 Answer 1

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For uniform convergence, notice that $f(x)=|x|=\sqrt{x^2}$.

Thus, $$f_n(x)-f(x)=\sqrt{x^2+1/n}-\sqrt{x^2}=\dfrac{1/n}{\sqrt{x^2+1/n}+\sqrt{x^2}}\leq \dfrac{1/n}{\sqrt{1/n}}=\dfrac{1}{\sqrt{n}}$$

Therefore, we can bound the difference with no dependence on the point $x$, and we have uniform convergence.

But $f(x)=|x|$ is not a differentiable function on $x=0$, which is a standard result.

Notice that this shows us that uniform convergence does not preserve differentiability.

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