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A sequence ($S_n$) of operators on a Banach space $X$ is said to be an approximate identity if $$ \|S_n f - f\|\rightarrow 0 \quad \forall f \in X $$

Could you please explain what the approximate identity is (logic behind the concept)?

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    A approximate identity is used in Fourier analysis. For example, if $f_n \to \delta$ (the Dirac delta function) then $S_n(f) := \delta * f$ is an approximate identity. – 2017-02-25
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    The way you have written it, an approximate identity is a sequence that converges to $\Bbb1$ in a weaker sense than norm topology. In this case you have written that it converges to $\Bbb1$ in the strong operator topology. – 2017-02-25

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For a more "analytic" answer than Munk's answer, which has a bit more of an "algebraic" character:

An approximate identity (in the sense that you've described) is a sequence of operators, usually derived from some "nice" class, that converge to the identity operator in the sense that you described. (Note that this sense is weaker than convergence in operator norm, which would imply uniformity over $f$.)

The classic example here is operators acting on some set of real-valued functions of a real variable, given by convolution against a smooth kernel. In this case the class itself contains no identity, but one can construct an approximate identity by taking a sequence of functions with integral $1$ which get "more and more concentrated" near $x=0$, such as Gaussian densities with variance decaying to zero.

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    Thank you for your reply! Is it possible to say that an approximate identity operator (for example, an integral operator that convolves a function with some kernel) makes the original function $f$ nicer in some sense while preserving a.e. convergence? http://msp.org/pjm/1979/81-1/pjm-v81-n1-p02-s.pdf – 2017-02-27
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    @Konstantin It doesn't necessarily make it nicer, but ones derived from convolution with a smooth kernel do, and generally speaking this is what you would want them for (if you don't want the additional "niceness", why would you want to *approximate* the identity at all?) – 2017-02-27
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My answer may be lacking since my main concern is C$^\ast$-algebras, but here is my point of view/intuition. For C$^\ast$-algebras (and Banach algebras) a unit might not exist. However, unital C$^\ast$-algebras are far easier to work with: techniques involving spectra for instance often come in handy. There are ways to "pass" non-unital problems to the unital realm eg. by regarding the unitization and/or multiplier algebra.

Using approximate units is another approach frequently used to "mimick" the unital case. The bonus here is that approximate units always exist (in fact, left ones always exists for left ideals) and using such bypasses the need to address an ambient unital algebra such as the unitization. A concrete example even exists in $B(H)$ for any Hilbert space $H$, say $H=\ell^2(\mathbb{N})$.

The compact operator operators $\mathbb{K}$ in $B(\ell^2(\mathbb{N}))$ contain a very concrete realization of an approximate unit. Indeed let $\lbrace \delta_n \rbrace_{n=1}^\infty$ be the canonical basis and let $p_n$ be the orthogonal projection onto the closed linear span of $\lbrace \delta_1, \ldots , \delta_n\rbrace$. Then $(p_n)_{n\geq 1}$ is an approximate unit of $\mathbb{K}$.

The above example is somewhat "the example" to me: $\mathbb{K}$ is non-unital, but the approximate looks more and more like the identity on $B(\ell^2(\mathbb{N}))$ as $n$ increases. As mentioned in the comments: $p_n$ converges in the strong-operator topology towards the identity (this is weaker than the norm topology!). To underline why approximate units may be preferred over regarding the unitization would include the following consequence:

  • Approximate units are used to verify that two-sided closed ideals of C$^\ast$-algebras are C$^\ast$-algebras themselves. Thus quotients of C$^\ast$-algebras are C$^\ast$-algebras as well. Approximate units are (as I recall) used to verify the C$^\ast$-identity, which cannot be done (as easily at least) using the unitization.